SequencesandSeries
Question1
Thenumberofcommontermsintheprogressions4,9,14,19,......,upto25th
termand3,6,9,12,upto37thtermis:
[27-Jan-2024Shift1]
Options:
A.
9
B.
5
C.
7
D.
8
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question2
[27-Jan-2024Shift1]
Answer:9
Solution:
-------------------------------------------------------------------------------------------------
Question3
The20thtermfromtheendoftheprogression
[27-Jan-2024Shift2]
Options:
A.
-118
B.
-110
C.
-115
D.
-100
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question4
IfinaG.P.of64terms,thesumofallthetermsis7timesthesumoftheodd
termsoftheG.P,thenthecommonratiooftheG.P.isequalto
[29-Jan-2024Shift1]
Options:
A.
7
B.
4
C.
5
D.
6
Answer:D
Solution:
-------------------------------------------------------------------------------------------------
Question5
InanA.P.,thesixthtermsa6=2.Ifthea1a4a5isthegreatest,thenthe
commondifferenceoftheA.P.,isequalto
[29-Jan-2024Shift1]
Options:
A.
3/2
B.
8/5
C.
2/3
D.
5/8
Answer:B
Solution:
-------------------------------------------------------------------------------------------------
Question6
Iflogea,logeb,logecareinanA.P.andlogea−loge2b,loge2b−loge3c,
loge3c−logeaarealsoinanA.P,thena:b:cisequalto
[29-Jan-2024Shift2]
Options:
A.
9:6:4
B.
16:4:1
C.
25:10:4
D.
6:3:2
Answer:A
Solution:
Question7
Ifeachtermofageometricprogressiona1,a2,a3,...witha1=1/8anda2≠
a1,isthearithmeticmeanofthenexttwotermsandSn=a1+a2+...+an,
thenS20−S18isequalto
[29-Jan-2024Shift2]
Options:
A.
215
B.
−218
C.
218
D.
−215
Answer:D
Solution:
Question8
LetSadenotethesumoffirstntermsanarithmeticprogression.IfS20=790
andS10=145,thenS15−S5is:
[30-Jan-2024Shift1]
Options:
A.
395
B.
390
C.
405
D.
410
Answer:A
Solution:
Question9
Letα=12+42+82+132+192+262+.......upto10termsand If4α
−β=55k+40,thenkisequalto
[30-Jan-2024Shift1]
Options:
Answer:353
Solution:
Question10
Letaandbbebetwodistinctpositiverealnumbers.Let11thtermofaGP,
whosefirsttermisaandthirdtermisb,isequaltopthtermofanotherGP,
whosefirsttermisaandfifthtermisb.Thenpisequalto
[30-Jan-2024Shift2]
Options:
A.
20
B.
25
C.
21
D.
24
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question11
LetSnbethesumton-termsofanarithmeticprogression3,7,11,.......
[30-Jan-2024Shift2]
Options:
Answer:9
Solution:
-------------------------------------------------------------------------------------------------
Question12
Thesumoftheseries upto10termsis
[31-Jan-2024Shift1]
Options:
A.
45/109
B.
C.
55/109
D.
Answer:D
Solution:
-------------------------------------------------------------------------------------------------
Question13
Let2nd,8thand44th,termsofanon-constantA.P.berespectivelythe1st,
2ndand3rdtermsofG.P.IfthefirsttermofA.P.is1thenthesumoffirst20
termsisequalto-
[31-Jan-2024Shift2]
Options:
A.
980
B.
960
C.
990
D.
970
Answer:D
Solution:
-------------------------------------------------------------------------------------------------
Question14
Let3,a,b,cbeinA.P.and3,a−1,b+1,c+9beinG.P.Then,thearithmetic
meanofa,bandcis:
[1-Feb-2024Shift1]
Options:
A.
-4
B.
-1
C.
13
D.
11
Answer:D
Solution:
-------------------------------------------------------------------------------------------------
Question15
Let3,7,11,15,....,403and2,5,8,11,..,404betwoarithmeticprogressions.
Thenthesum,ofthecommontermsinthem,isequalto____
[1-Feb-2024Shift1]
Options:
Answer:6699
Solution:
-------------------------------------------------------------------------------------------------
Question16
LetSndenotethesumofthefirstntermsofanarithmeticprogression.
IfS10=390andtheratioofthetenthandthefifthtermsis15:7,thenS15−
S5isequalto:
[1-Feb-2024Shift2]
Options:
A.
800
B.
890
C.
790
D.
690
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question17
IfthreesuccessivetermsofaG.P.withcommonratior(r>1)arethelengths
ofthesidesofatriangleand[r]denotesthegreatestintegerlessthanor
equaltor,then3[r]+[−r]isequalto:
[1-Feb-2024Shift2]
Options:
Answer:1
Solution:
Question18
Forthreepositiveintegersp,q,r,xpq q2=yqr =zp2randr =pq +1suchthat
3,3logyx,3logzy,7logxzareinA.P.withcommondifference 1
2.Thenr pq
isequalto
[24-Jan-2023Shift1]
Options:
A.2
B.6
C.12
D.6
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question19
The4 th termofGPis500anditscommonratiois 1
m,mN.LetSndenote
thesumofthefirstntermsofthisGP.IfS6>S5+1andS7<S6+1
2,thenthe
numberofpossiblevaluesofmis
[24-Jan-2023Shift1]
Answer:12
Solution:
pq2=logxλ
qr =logyλ
p2r=logzλ
logyx=qr
pq2=r
pq... . (1)
logxz=pq2
p2r=q2
pr..........(2)
logzy=p2r
qr =p2
q..... . (3)
3,3r
pq,3p2
q,7q2
pr inA.P
3r
pq 3=1
2
r=7
6pq......(4)
r=pq +1
pq =6......(5)
r=7..... . (6)
3p2
q=4
Aftersolvingp=2andq=3
-------------------------------------------------------------------------------------------------
Question20
If 13+23+33+ ...... upto n terms
13+25+37+ ...... upto n terms =9
5,thenthevalueofnis
[24-Jan-2023Shift2]
Answer:5
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question21
LetA1,A2,A3bethethreeA.P.withthesamecommondifferencedand
havingtheirfirsttermsasA,A+1,A+2,respectively.Leta,b,cbethe
T4=500 where a =firstterm,
r=commonratio =1
m,mN
ar3=500
a
m3=500
SnSn1=arn1
S6>S5+1 and S7S6<1
2
S6S5>1a
m6<1
2
ar5>1 m3>103
500
m2>1 m >10
m2<500 ......(1)
From(1)and(2)
m=11,12,13............., 22
Sonumberofpossiblevaluesofmis12
13+23+33..... + n3=n(n+1)
2
2
13+25+37+ ... + nn+terms =
n
r=1
r(2r +1) =
n
r=1
(2r2+r)
=2n(n+1)(2n +1)
6+n(n+1)
2
=n(n+1)
6(2(2n +1) + 3)
=n(n+1)
2×(4n +5)
3
=n2(n+1)2
4
5n(n+1)
2×(4n +5)
3
2=9
5
5n(n+1)
2=9(4n +5)
3
15n(n+1) = 18(4n +5)
15n2+15n =72n +90
15n257n 90 =05n219n 30 =0
(n5)(5n +6) = 0
n=6
5or 5
n=5.
( )
7th ,9th ,17 th termsofA1,A2,A3,respectivelysuchthat
a 7 1
2b 17 1
c 17 1
+70 =0
Ifa =29,thenthesumoffirst20termsofanAPwhosefirsttermisc ab
andcommondifferenceis d
12,isequalto________.
[25-Jan-2023Shift1]
Answer:495
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question22
Forthetwopositivenumbersa,b,ifa,band 1
18areinageometric
progression,while 1
a,10and 1
bareinanarithmeticprogression,then,
16a +12bisequalto________.
[25-Jan-2023Shift2]
Answer:3
Solution:
Solution:
| |
A+6d 7 1
2(A+1+8d)17 1
A+2+16d 17 1
+70 =0
A= 7 and d =6
cab=20
S20 =495
| |
a,b,1
18 GP
a
18 =b2
1
a,10,1
bAP
1
a+1
b=20
a+b=20 ab,fromeq.(i) ;weget
18b2+b=360b3
360b218b 1=0{∵b0}
b=18 ± 324 +1440
720
b=18 + 1764
720 {∵b>0}
b=1
12
⇒=18×1
144 =1
8
-------------------------------------------------------------------------------------------------
Question23
Leta1,a2,a3, ....beaGPofincreasingpositivenumbers.Iftheproductof
fourthandsixthtermsis9andthesumoffifthandseventhtermsis24,then
a1a9+a2a4a9+a5+a7isequalto______.
[29-Jan-2023Shift1]
Answer:60
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question24
Let{ak}and{bk}, kN,betwoG.P.swithcommonratior1andr2
respectivelysuchthata1=b1=4andr1<r2.Letck=ak+bk,kN.Ifc2=5
andc3=13
4then
k=1ck (12a6+8b4)isequalto__
[29-Jan-2023Shift2]
Answer:9
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question25
Now,16a +12b =16×1
8+12×1
12 =3
a4a6=9 (a5)2=9a5=3
&a5+a7=24 a5+a5r2=24 (1+r2) = 8r= 7
a=3
49
a1a9+a2a4a9+a5+a7=9+27 +3+21 =60
Giventhat
ck=ak+bkand a1=b1=4
also a2=4r1a3=4r1
2
b2=4r2b3=4r2
2
Nowc2=a2+b2=5andc3=a3+b3=13
4
r1+r2=5
4andr1
2+r2
2=13
16
Hencer1r2=3
8whichgivesr1=1
2&r2=3
4
k
k=1c
(12a6+8b4)
=4
1r1
+4
1r2
48
32 +27
2
=24 15 =9
( )
Leta1=b1=1andan=an1+ (n1), bn=bn1+an1, n2.IfS =
n=110 bn
n
2
andT =
n=18n
n1
2
,then27(2S T)isequalto______.
[29-Jan-2023Shift2]
Answer:461
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question26
Ifan=2
4n216n +15,thena1+a2+ ...... + a25isequalto:
[30-Jan-2023Shift1]
Options:
A. 51
144
B. 49
138
C. 50
141
D. 52
147
As, S =b1
2+b2
22+ ...... + b9
29+b10
210
S
2=b1
22+b2
23+ ...... + b9
210 +b10
211
subtracting
S
2=b1
2+a1
22+a2
23...... + a9
210 b10
211
S=b1b10
210 +a1
2+a2
22........ + a9
29
S
2=b1
2b10
211 +a1
22+a2
23....... + a9
210
subtracting
s
2=b1
2b10
211 +a1
2a9
210 +1
22+2
23+ ... + 8
29
S
2=a1+b1
2(b10 +2a9)
211 +T
4
2S =2(a1+b1) (b10 +2a9)
29+T
27(2S T) = 28(a1+b1) (b10 +2a9)
4
Also, bnbn1=an1
b10 b1=a1+a2+ ... + a9
=1+2+4+7+11 +16 +22 +29 +37
b10 =130(As b1=1)
27(2S T) = 28(1+1) (130 +2×37)
29204
4=461
( )
( )
( )
( ) ( )
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question27
Let
n=0
n3((2n)!) + (2n 1)(n!)
(n!)((2n)!) =ae +b
e+c,wherea,b,c ande =
n=0
1
nThen
a2b+cisequalto_______.
[30-Jan-2023Shift1]
Answer:26
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question28
Leta,b,c>1,a3,b3andc3beinA.P.,andlogab,logcaandlogbcbeinG.P.If
thesumoffirst20termsofanA.P.,whosefirsttermis a+4b +c
3andthe
commondifferenceis a8b +c
10 is444,thenabcisequalto
[30-Jan-2023Shift2]
Options:
A.343
Option(3)
Ifan=2
4n216n +15thena1+a2+ ........ . a25
25
n=1
an= 2
4n216n +15
= 2
4n26n 10n +15
= 2
2n(2n 3) 5(2n 3)
= 2
(2n 3)(2n 5)
= 1
2n 31
2n 5
=1
47 1
(−3)
=50
141
n=0
n3((2n)!) + (2n 1)(n!)
(n!)((2n)!)
=
n=0
1
(n3)! +
n=0
3
(n2)!
+
n=0
1
(n1)! +
n=0
1
(2n 1)!
n=0
1
(2n)!
=e+3e +e+1
2e1
e1
2e+1
e
=5e 1
e
a2b+c=26
( ) ( )
B.216
C. 343
8
D. 125
8
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question29
Theparabolas:ax2+2bx +cy =0andd x2+2ex +f y =0intersectontheline
y=1.Ifa,b,c,d,e,f arepositiverealnumbersanda,b,careinG.P.,then
[30-Jan-2023Shift2]
Options:
A.d,e,fareinA.P.
B. d
a,e
b,f
careinG.P.
C. d
a,e
b,f
careinA.P.
D.d,e,fareinG.P.
Answer:C
Solution:
Solution:
Asa3,b3,c3beinA.P.→a3+c3=2b3...(1)
loga
b,logc
a,logb
careinG.P.
log b
log a log c
log b =log a
log c
2
(log a)3= (log c)3a=c...(2)
From(1)and(2)
a=b=c
T1=a+4b +c
3=2a;d=a8b +c
10 =6a
10 =3
5a
S20 =20
24a +19 3
5a
=10 20a 57a
5
= 74a
74a = 444 a=6
abc =63=216
( )
[ ( ) ]
[ ]
ax2+2bx +c=0
ax2+2acx +c=0(∵b2=ac)
(xa+ c)2=0
x2c
a......
Now, dx2+2ex +f=0
dc
a+2e c
a+f=0
d c
a+f=2e c
a
d
a+f
c=2e 1
ac
d
a+f
c=2e
b[as b = ae]
( ) [ ]
-------------------------------------------------------------------------------------------------
Question30
The8 th commontermoftheseries
S1=3+7+11 +15 +19 + ....
S2=1+6+11 +16 +21 + ...
is_______.
[30-Jan-2023Shift2]
Answer:151
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question31
Lety =f(x)representaparabolawithfocus 1
2,0 anddirectrixy = 1
2.
Then
S=xR:tan1(√f(x)+sin 1(√f(x) + 1)) = π
2:
[31-Jan-2023Shift1]
Options:
A.containsexactlytwoelements
B.containsexactlyoneelement
C.isaninfiniteset
D.isanemptyset
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question32
( )
{ }
d
a,e
b,f
careinA.P.
T8=11 + (81) × 20
=11 +140 =151
x+1
2
2=y+1
4
y= (x2+x)
tan1x(x+1)+sin 1x2+x+1=π2
0x2+x+11
x2+x0...(1)
Also x2+x0...(2)
x2+x=0x=0, 1
Scontains2element.
( ) ( )
Leta1,a2, ......, anbeinA.P.Ifa5=2a7anda11 =18,then
12 1
a10 + a11
+1
a11 + a12
+ .... 1
a17 + a18
isequalto_______.
[31-Jan-2023Shift1]
Answer:8
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question33
Leta1,a2,a3, ....beanA.P.Ifa7=3,theproducta1a4isminimumandthe
sumofitsfirstntermsiszero,thenn! 4an(n+2)isequalto:
[31-Jan-2023Shift2]
Options:
A.24
B. 33
4
C. 381
4
D.9
Answer:A
Solution:
Solution:
( )
2a7=a5(given)
2(a1+6d ) = a1+4d
a1+8d =0...(1)
a1+10d =18...(2)
By (1)and (2)weget a1= 72,d=9
a18 =a1+17d = 72 +153 =81
a10 =a1+9d =9
12 a11 a10
d+a12 a11
d+ ...... a18 a17
d
12 a18 a10
d=12(93)
9=12 ×6
6=8
( )
( )
a+6d =3...(1)
Z=a(a+3d )
= (36d )(33d )
=18d 227d +9
Differentiatingwithrespecttod
36d 27 =0
d=3
4,from(1)a=3
2, ( Z= minimum)
Now,Sn=n
23+ (n1)3
4=0
n=5
Now,
n! 4an(n+2)=120 4(a35)
=120 4(a+ (35 1)d)
( )
-------------------------------------------------------------------------------------------------
Question34
Thesum
122.32+3.524.72+5.92 ........ + 15.292is________.
[31-Jan-2023Shift2]
Answer:6952
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question35
Thesumto10termsoftheseries 1
1+12+14+2
1+22+24+3
1+32+34+ ....is:-
[1-Feb-2023Shift1]
Options:
A. 59
111
B. 55
111
C. 56
111
D. 58
111
Answer:B
Solution:
Solution:
=120 43
2+34 3
4
=120 46+102
4
=120 96 =24
( ( ) )
( )
Separatingoddplacedandevenplacedtermsweget
S= (1.12+3.52+ .. . .15 (29)2) ( 2.32+4.72.
+ ... + 14 . (27)2
S=
8
n=1
(2n 1)(4n 3)2
7
n=1
(2n)(4n 1)2
Applyingsummationformulaweget
=29856 22904 =6952
Tr=(r2+r+1) (r2r+1)
2(r4+r2+1)
Tr=1
2
1
r2r+11
r2+r+1
T1=1
2
1
11
3
T2=1
2
1
31
7
T3=1
2
1
71
13
[ ]
[ ]
[ ]
[ ]
-------------------------------------------------------------------------------------------------
Question36
Leta1=8,a2,a3, ... . anbeanA.P.Ifthesumofitsfirstfourtermsis50and
thesumofitslastfourtermsis170,thentheproductofitsmiddletwo
termsis______.
[1-Feb-2023Shift1]
Answer:754
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question37
Thenumberof3-digitnumbers,thataredivisiblebyeither2or3butnot
divisibleby7is_______.
[1-Feb-2023Shift1]
Answer:514
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question38
Whichofthefollowingstatementsisatautology?
[1-Feb-2023Shift2]
Options:
T10 =1
2
1
91 1
111
10
r=1
Tr=1
211
111 =55
111
[ ]
[ ]
a1+a2+a3+a4=50
32 +6d =50
d=3
and, an3+an2+an1+an=170
32 + (4n 10) 3=170
n=14
a7=26,a8=29
a7a8=754
Divisibleby2450
Divisibleby3300
Divisibleby7128
Divisibleby2&764
Divisibleby3&743
Divisibleby2&3150
Divisibleby2,3&721
∴Totalnumbers = 450 +300 150 64 43 +21 =514
A.p ((pq))
B.(pΛq) (∼(p) q) )
C.((pq)) q
D.pV(pΛq)
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question39
Thesumofthecommontermsofthefollowingthreearithmeticprogressions.
3,7,11,15, ............, 399
2,5,8,11, ............., 359and
2,7,12,17, ......, 197,isequalto________.
[1-Feb-2023Shift2]
Answer:321
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question40
Thesumofthefirst20termsoftheseries5 +11 +19 +29 +41 + ......is:
[6-Apr-2023shift1]
Options:
A.3450
B.3420
(i)p ((pq))
(∼p)V((∼pVq))
(∼p)V(fV(pΛq))
pV(pΛq) = (∼pVp)Λ(∼pVq)
= pVq
(ii)(pΛq) (∼pq)
(pΛq)V(pq) = t
{a,b,d}V{a,b,c} = V
Tautology
(iii)((pq)) q
((∼pVq))Vq= (pΛq)Vq= pV q
Nottantology
(iv)pV (pΛq) = p
Nottautology.
3,7,11,15, ............, 399 d1=4
2,5,8,11, ............., 359 d2=3
2,7,12,17, ......, 197 d3=5
LCM(d1,d2,d3) = 60
Commontermsare47,107, 167
Sum =321
C.3520
D.3250
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question41
Leta1,a2,a3, ..., anbenpositiveconsecutivetermsofanarithmetic
progression.Ifd >0isitscommondifference,then:
lim
n
d
n
1
a1+ a2
+1
a2+ a3
+ ....... + 1
an1+ an
is
[6-Apr-2023shift1]
Options:
A. 1
d
B.1
C.d
D.0
Answer:B
Solution:
Solution:
( )
Tn=5+(n1)
2[26+ (n2) 2]
Tn=5+ (n1)(n+4) = 5+n2+3n 4=n2+3n +1
Now S20 =
20
n=1
Tn=
20
n=1
n2+3n +1
S20 =20.21.41
6+3.20.21
2+20
S20 =2870 +630 +20
S20 =3520
Lt
n
d
n
a1 a2
a1a2
+a2 a3
a2a3
+ ...... + an1 an
an1an
=Lt
n
d
n
a1 a2+ a2+ a3+ ...... + an1 an
d
=Lt
n
d
n
an a1
d
=Lt
n
1
n
a1+ (n1)d a1
d
=Lt
n
1
d
a1
n+dd
na1
n
=1
( )
( )
( )
( )
()
-------------------------------------------------------------------------------------------------
Question42
Ifgcd(m,n) = 1and
1222+3242+ ...... + (2021)2 (2022)2+ (2023)2=1012m2n
thenm2n2isequalto:
[6-Apr-2023shift2]
Options:
A.180
B.220
C.200
D.240
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question43
If(20)19 +2(21)(20)18 +3(21)2(20)17 + ... + 20(21)19 =k(20)19,thenkisequal
to________:
[6-Apr-2023shift2]
Answer:400
Solution:
Solution:
(12)(1+2) + (34)(3+4) + ...... + (2021 2022)(2021 +2022) + (2023)2= (1012)m2n
(−1)[1+2+3+4+ ... + 2022] + (2023)2= (1012)m2n
(−1)(2022)(2023)
2+ (2023)2= (1012)m2n
(2023)[2023 1011] = (1012)m2n
(2023)(1012) = (1012)m2n
m2n=2023
m2n= (17)2×7
m=17,n=7
m2n2= (17)272=289 49 =240
Ans.Option4
S= (20)19 +2(21)(20)18 + ...... + 20(21)19
21
20S=21(20)18 +2(21)9(20)17 + ....... + (21)20
Subtract
121
20 S= (20)19 + (21)(20)18 + (21)2(20)17 + ...... + (21)19 (21)20
1
20 S= (20)19 121
20
20
121
20
(21)20
1
20 S= (21)20 (20)20 (21)20
S= (20)21 =K(20)19 (given)
K= (20)2
=400
( )
( ) [( ) ]
( )
Question44
LetSK
1+2+ ... + K
Kand n
j=1Sj
2=n
A(Bn2+Cn +D),whereA,B,C,DNandAhas
leastvalue.Then
[8-Apr-2023shift1]
Options:
A.A +BisdivisiblebyD
B.A +B=5(DC)
C.A +C+DisnotdivisiblebyB
D.A +B+Disdivisibleby5
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question45
Letanbethen th termoftheseries5 +8+14 +23 +35 +50 + ....and
Sn =n
k=1aK.ThenS30 a40isequalto
[8-Apr-2023shift2]
Options:
A.11260
B.11280
C.11290
D.11310
Answer:C
Solution:
Solution:
Sk=k+1
2
Sk
2=k2+1+2k
4
n
j1
Sj
2=1
4
n(n+1)(2n +1)
6+n+n(n+1)
=n
4
(n+1)(2n +1)
6+1+n+1
=n
4
2n2+3n +1
6+n+2
=n
4
2n2+9n +13
6=n
24[2n2+9n +13]
A=24,B=2,C=9,D=13
[ ]
[ ]
[ ]
[ ]
Sn=5+8+14 +23 +35 +50 + ... + an
Sn=5+8+14 +23 +35 + ... + an
────────────────────────────
O=5+3+6+9+12 +15 + ... an
-------------------------------------------------------------------------------------------------
Question46
Let0 <z<y<xbethreerealnumberssuchthat 1
x,1
y,1
zareinanarithmetic
progressionandx, 2y,zareinageometricprogression.If
xy +yz +zx =3
2xyz,then3(x+y+z)2isequalto_______.
[8-Apr-2023shift2]
Answer:150
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question47
Letthefirsttermaandthecommonratiorofageometricprogressionbe
positiveintegers.Ifthesumofsquaresofitsfirstthreeis33033,thenthe
sumofthesetermsisequalto:
[10-Apr-2023shift1]
Options:
A.210
B.220
an=5+ (3+6+9+ .. . (n1)terms )
an=3n23n +10
2
a40 =3(40)23(40) + 10
2=2345
S30 =
3
30
Σ
n=1
n23
30
Σ
n=1
n+10
30
Σ
n=1
1
2
=
3×30 ×31 ×61
63×30 ×31
2+10 ×30
2
S30 =13635
S30 a40 =13635 2345
=11290(Option (3))
2
y=1
x+1
z
2y2=xz
2
y=x+z
xz =x+z
2y2
x+z=4y
xy +yz +zx =3
2xyz
y(x+z) + zx =3
2xz y
4y2+2y2=3
2y2y2
6y2=32y3
y= 2
x+y+z=5y =52
3(x+y+z)2=3×50 =150
C.231
D.241
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question48
Iff (x) = (tan 1)x+loge(123)
xloge(1234) (tan 1),x>0,thentheleastvalueoff (f(x)) + f f 4
xis:
[10-Apr-2023shift1]
Options:
A.2
B.4
C.8
D.0
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question49
( ( ) )
Leta,ar,ar2bethreetermsofGP
Given:a2+ (ar)2+ (ar2)2=33033
a2(1+r2+r4) = 112.3.7.13
a=11 and 1 +r2+r4=3.7.13
r2(1+r2) = 273 1
r2(r2+1) = 272 =16 ×17
r2=16
r=4[∵r>0]
Sumofthreeterms = a+ar +ar2=a(1+r+r2)
=11(1+4+16)
=11 ×21 =231
f(x) = (tan 1)x+log 123
x log 1234 tan 1
LetA=tan 1,B=log 123,C=log 1234
f(x) = Ax +B
xC A
f(f(x)) =
AAx +B
xC A+B
CAx +B
CX AA
=A2x+AB +xBC AB
ACx +BC ACx +A2
=x(A2+BC)
(A2+BC.=x
f(f(x)) = x
f f 4
x=4
x
f(f(x)) + f f 4
x
AM GM
x+4
x4
( )
( )
( ( ) )
( ( ) )
Thesumofallthoseterms,ofthearithmeticprogression3,8,13, ...., 373,
whicharenotdivisibleby3,isequalto_______.
[10-Apr-2023shift1]
Answer:9525
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question50
IfSn=4+11 +21 +34 +50 + ....tonterms,then 1
60(S29 S9)isequalto
[10-Apr-2023shift2]
Options:
A.220
B.227
C.226
D.223
Answer:D
Solution:
Solution:
A.P:3,8,13... . .373
Tn=a+ (n1)d
373 =3+ (n1)5
n=370
5
n=75
NowSum =n
2[a+1]
=75
2[3+373] = 14100
Nownumbersdivisibleby3are,
3,18,33..... . .363
363 =3+ (k1)15
k1=360
15 =24 k=25
Now,sum =25
2(3+363) = 4575s
req.sum =14100 4575
=9525
Sn=4+11 +21 +34 +50 + ... + nterms
DifferenceareinA.P.
LetTn=an2+bn +c
T1=a+b+c=4
T2=4a +2b +c=11
T3=9a +3b +c=21
Bysolvingthese3equations
a=3
2,b=5
2,c=0
SoTn=3
2n2+5
2n
Sn=ΣT n
=3
2n2+5
2n
-------------------------------------------------------------------------------------------------
Question51
Supposea1,a2,2,a3,a4beinanarithemetico-geometricprogression.Ifthe
commonratioofthecorrespondinggeometricprogressionin2andthesum
ofall5termsofthearithmetico-geometricprogressionis 49
2,thena4isequal
to_______.
[10-Apr-2023shift2]
Answer:16
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question52
Letx1,x2, ....., x100beinanarithmeticprogression,withx1=2andtheir
meanequalto200.Ifyi=i(xii),1 i100,thenthemeanof
y1,y2, .... . y100is:
[11-Apr-2023shift1]
Options:
A.10051.50
B.10100
C.10101.50
D.10049.50
Answer:D
Solution:
Solution:
=3
2
n(n+1)(2n +1)
6+5
2
(n)(n+1)
2
=n(n+1)
4[2n +1+5]
Sn=n(n+1)
4(2n +6) = n(n+1)(n+3)
2
1
60
29 ×30 ×32
29×10 ×12
2=223
( )
(a2d )
4,(ad)
2,a,2(a+d), 4(a+2d )
a=2
1
4+1
2+1+6×2+ (−1+2+8)d=49
2
23
4+7+9d =49
2
9d =49
262
4=98 62
4=9
d=1
a4=4(a+2d )
=16
( )
( )
-------------------------------------------------------------------------------------------------
Question53
LetS =S=109 +108
5+107
52+ .... + 2
5107 +1
5108.Thenthevalueof(16S (25)54)
isequalto_______.
[11-Apr-2023shift1]
Answer:2175
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question54
Leta,b,canddbepositiverealnumberssuchthata +b+c+d=11.Ifthe
maximumvalueofa5b3c2dis3750β,thenthevalueofβis
[11-Apr-2023shift2]
Options:
A.55
Mean =200
100
2(2×2+99d)
100 =200
4+99d =400
d=4
yi=i(xi 1)
=i(2+ (i1)4i) = 3i22i
Mean =Σyi
100
=1
100
100
i=1
3i22i
=1
100
3×100 ×101 ×201
62×100 ×101
2
=101 201
21=101 ×99.5
=10049.50
{ }
{ }
S=109 +108
5+107
52...... + 1
5108
S
5=109
5+108
52...... 2
5108 +1
5109
4S
5=109 1
51
52...... 1
5108 1
5109
=109 1
5
11
5109
11
5
=109 1
411
5109
=109 1
4+1
4×1
5109
S=5
4109 1
4+1
4.5109
16S =20 ×109 5+1
5108
16S (25)54 =2180 5=2175
(( ) )
( ) )
( )
( )
B.108
C.90
D.110
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question55
Fork N,ifthesumoftheseries1 +4
k+8
k2+13
k3+19
k4+ ......is10,thenthe
valueofkis______
[11-Apr-2023shift2]
Answer:2
Solution:
Solution:
Givena+b+c+d=11
(a,b,c,d>0}
(a5b3c2d)max.=?
LetassumeNumbers-
a
5,a
5,a
5,a
5,a
5,b
3,b
3,b
3,c
2,c
2,
WeknowA.M.≥G.M.
a
5+a
5+a
5+a
5+a
5+b
3+b
3+b
3+c
2+c
2+d
11 a5b3c2d
5533221
1
11
11
11 a5b3c2d
5533221
1
11
a5b3c2d553322
max(a5b3c2d) = 553322=337500
=90 ×3750 =β×3750
β=90
Option(C)90correct
( )
( )
10 =1+4
k+8
k2+13
k3+19
k4+ ...... .upto
9=4
k+8
k2+13
k3+19
k4+ ....... upto
9
k=4
k2+8
k3+13
k4+ ..... .upto
_________________________________
S=9 1 1
k=4
k+4
k2+5
k3+6
k4...... upto
S
k=4
k2+4
k3+5
k4+ ..... .upto
__________________________________
11
kS=4
k+1
k3+1
k4+1
k5+ .... .
9 1 1
k
2=4
k+
1
k3
11
k
9(k1)3=4k(k1) + 1
k=2
( )
( )
( ) ( )
-------------------------------------------------------------------------------------------------
Question56
Let<an>beasequencesuchthata1+a2+ .. + an=n2+3n
(n+1)(n+2).If
28 10
k=1
1
ak
=p1p2p3...pm,wherep1,p2......pmarethefirstmprimenumbers,
thenmisequalto
[12-Apr-2023shift1]
Options:
A.8
B.5
C.6
D.7
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question57
Lets1,s2,s3, ......., s10respectivelybethesumto12termsof10A.P.sm
whosefirsttermsare1,2,3, ..., 10andthecommondifferencesare
1,3,5, .........., 19respectively.Then 10
i=1siisequalto
[13-Apr-2023shift1]
Options:
A.7260
B.7380
C.7220
D.7360
Answer:A
Solution:
Solution:
an=SnSn1=n2+3n
(n+1)(1+2)(n1)(n+2)
n(n+1)
an=4
n(n+1)(1+2)
28
10
k1
1
ak
=28
10
k1
k(k+1)(k+2)
4
=7
4
10
k1
(k(k+1)(k+2)(k+3) (k1)k(k+1)(k+2))
=7
410 11 12 13 =235711 13
Som=6
Sk=6(2k + (11)(2k 1))
Sk=6(2k +22k 11)
Sk=144k 66
-------------------------------------------------------------------------------------------------
Question58
Thesumto20termsoftheseries2.2232+2.4252+2.62 ........isequal
to_______.
[13-Apr-2023shift1]
Answer:1310
Solution:
-------------------------------------------------------------------------------------------------
Question59
Leta1,a2,a3, ....beaG.P.ofincreasingpositivenumbers.Letthesumofits
6th and8 th termsbe2andtheproductofits3rdand5 th termsbe 1
9.Then
6(a2+a4)(a4+a6)isequalto
[13-Apr-2023shift2]
Options:
A.2
B.3
C.33
D.22
Answer:B
Solution:
Solution:
10
1
Sk=144
10
k=1
k66 ×10
=144×10 ×11
2660
=7920 660
=7260
(2232+4252+20 terms ) + (22+42+ .... + 10 terms )
(2+3+4+5+ .... + 11) + 4[1+22+ ... . .102]
21 ×22
21+4×10 ×11 ×21
6
=1231 +14 ×11 ×10
=1540 +1231
=1310
[ ]
a3a5=1
9
ar2ar4=1
9
(ar3)2=1
9
ar3=1
3...(i)
a6+a8=2
ar5+ar7=2
-------------------------------------------------------------------------------------------------
Question60
Let[α]denotethegreatestintegerα.Then[√1] + [√2] + [√3] + ... + [√120]
isequalto_______.
[13-Apr-2023shift2]
Answer:825
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question61
Letf (x) = 10
k=1kxk,xRIf2f ′(2) f(2) = 119(2)n+1thennisequalto_______
[13-Apr-2023shift2]
Answer:10
Solution:
ar3(r2+r4) = 2
1
3r2(1+r2) = 2
r2(1+r2) = 2×3
r2=2r= 2
a=1
3×1
r3
=1
3×1
22=1
62
6(a2+a4)(a4+a6)
6(ar +ar3)(ar3+ar5)
6ar3
r2+1
3
1
3+1
3r2=3
( ) ( )
S= [√1] + [√2] + [√3] + ... + [√120]
[√1] [√3] = 1×3
[√4] [√8] = 2×5
[√9] [√15] = 3×7
[√100] [√120] = 10 ×21
S=1×3+2×5+3×7+ ... + 10 ×21
=
10
r=1
r(2r +1)
=2
10
r=1
r2+
10
r=1
r
=2×10 ×11 ×21
6+10 ×11
2
=770 +55
=825
f(x) =
10
k=1
kx xk
f(x) = x+2x2+3x3+ +9x9+10x10 (i)
-------------------------------------------------------------------------------------------------
Question62
LetA1andA2betwoarithmeticmeansandG1,G2,G3bethreegeometric
meansoftwodistinctpositivenumbers.ThenG1
4+G2
4+G3
4+G1
2G3
2is
equalto
[15-Apr-2023shift1]
Options:
A.2(A1+A2)G1G3
B.(A1+A2)2G1G3
C.2(A1+A2)G1
2G3
2
D.(A1+A2)G1
2G3
2
Answer:B
Solution:
Solution:
xf(x) = x2+2x3+ ... + 9x10 +10x11... (ii)
"(i) (ii)1
f(x)(1x) = x+x2+x3+ +x10 10x11
f(x)(1x) = x(1x10)
1x10x11
f(x) = x(1x10)
(1x)210x11
(1x)
f(2) = 2+g(2)11
(1x)2f(x) = x(1x10) 10x11(1x)
diff.w.r.t. x
(1x)2f(2) + f(2)2(1x)(−1)
=x(−10x9) + (1x10) 10x11(−1) (1x)(110)x10
put x =2
f(2) + f(2)(2) = 10(2)10 +1210 +10(2)11 110(2)10 +110(2)11
= (−121)210 + (120)211 +1
=210(240 121) + 1
=119(2)10 +1
n=10
a,A1,A2,bareinA.P.
d=ba
3;A1=a+ba
3=2a +b
3
A2=a+2b
3
A1+A2=a+b
a,G1,G2,G3,bareinG.P.
r=b
a
1
4
G1= (a3b)
1
4
G2= (a2b2)
1
4
G3= (ab3)
1
4
G1
4+2
4+G3
4+G1
2G3
2=
a3b+a2b2+ab3+ (a3b)
1
2 (ab3)
1
2
=a3b+a2b2+ab3+a2b2
=ab(a2+2ab +b2)
=ab(a+b)2
( )
-------------------------------------------------------------------------------------------------
Question63
Ifthesumoftheseries 1
21
3+1
221
23+1
32+1
231
223+1
2321
33+
1
241
223+1
22321
232+1
34+ .......is α
β,whereαandβareco-prime,then
α+3βisequalto_______
[15-Apr-2023shift1]
Answer:7
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question64
If{ai}n
i=1,wherenisaneveninteger,isanarithmeticprogressionwith
commondifference
1,and n
i=1ai=192,n2
i=1a2i =120,thennisequalto:
[24-Jun-2022-Shift-1]
Options:
A.48
B.96
C.92
D.104
Answer:B
Solution:
( ) ( ) ( )
( )
=G1G3 (A1+A2)2
P1
21
3+1
221
23+1
32+1
23+1
223+1
2321
33+ .. . P1
2+1
3=1
221
32+1
23+1
33+1
241
34+ ...
5P
6=
1
4
11
2
1
9
1+1
3
5P
6=1
21
12 =5
12
P=1
2=α
βα=1,β=2
α+ =7
( ) ( ) ( ) ( ) ( ) ( ) ( )
Question65
If 1
2310 +1
2239+ ... + 1
210 3=K
210 310,thentheremainderwhenKisdividedby6is
:
[25-Jun-2022-Shift-1]
Options:
------------------------------------------------------------------------------------------------
A.1
B.2
C.3
D.5
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question66
Thegreatestintegerlessthanorequaltothesumoffirst100termsofthe
sequence 1
3,5
9,19
27,65
81,.isequalto
[25-Jun-2022-Shift-1]
Options:
A.
Answer:98
Solution:
-------------------------------------------------------------------------------------------------
Question67
Thesum1 +23+332+ ......... + 10.39isequalto
[25-Jun-2022-Shift-2]
Options:
A. 2312 +10
4
B. 19 310 +1
4
C.5 310 2
D. 9310 +1
2
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question68
LetA =10
i=1
10
j=1min{i,j}andB =10
i=1
10
j=1max{i,j}.ThenA +Bisequalto
[26-Jun-2022-Shift-1]
Answer:1100
Solution:
LetS=1.30+2.31+3.32+ ...... + 10.39
3S =1.31+2.32+ ......... + 10.310
2S = (1.30+1.31+1.32+ ...... + 1.39) 10.310
S=1
210.310 310 1
31
S=19.310 +1
4
[ ]
-------------------------------------------------------------------------------------------------
Question69
IfA =
n=1
1
(3+ (−1)n)nandB =
n=1
(−1)n
(3+ (−1)n)n,then A
Bisequalto:
[26-Jun-2022-Shift-2]
Options:
A. 11
9
B.1
C.11
9
D.11
3
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question70
Ifa1(>0), a2,a3,a4,a5areinaG.P.,a2+a4=2a3+1and3a2+a3=2a4,then
a2+a4+2a5isequalto____
[26-Jun-2022-Shift-2]
Answer:40
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question71
x=
n=0an,y=
n=0bn,z=
n=0cn,wherea,b,careinA.P.and
|a| < 1, | b| <1, | c| <1,abc0,then:
[27-Jun-2022-Shift-1]
Options:
A.x,y,zareinA.P.
B.x,y,zareinG.P.
C. 1
x,1
y,1
zareinA.P.
D. 1
x+1
y+1
z=1 (a+b+c)
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question72
Ifthesumofthefirsttentermsoftheseries
1
5+2
65 +3
325 +4
1025 +5
2501 + ...
is m
n,wheremandnareco-primenumbers,thenm +nisequalto
[27-Jun-2022-Shift-1]
Answer:276
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question73
LetS =2+6
7+12
72+20
73+30
74+ .....Then4Sisequalto
[27-Jun-2022-Shift-2]
Options:
A. 7
3
2
B. 73
32
C. 7
3
3
D. 72
33
Answer:C
( )
( )
x=
n=0
an=1
1a;y=
n=0
bn=1
1b;z=
n=0
cn=1
1c
Now,
a,b,cAP
1a,1b,1cAP
1
1a,1
1b,1
1cH P
x,y,zHP
1
x,1
y,1
zAP
Tr=r
(2r2)2+1
=r
(2r2+1)2 (2r)2
=1
4
4r
(2r2+2r +1)(2r22r +1)
S10 =1
4
10
r=1
1
(2r22r +1)1
(2r2+2r +1)
=1
411
5+1
51
13 + ... + 1
181 1
221
S10 =1
4220
221 =55
221 =m
n
m+n=276
( )
[ ]
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question74
Ifa1,a2,a3......andb1,b2,b3......areA.P.,and
a1=2,a10 =3,a1b1=1=a10b10,thena4b4isequalto-
[27-Jun-2022-Shift-2]
Options:
A. 35
27
B.1
C. 27
28
D. 28
27
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
S=2+6
7+12
72+20
73+30
74+ .........(i)
1
7S=2
7+6
72+12
73+20
74+ .........(ii)
(i)-(ii)
6
7S=2+4
7+6
72+8
73+ ............(iii)
6
72S=2
7+4
72+6
73+ ............(iv)
(iii)-(iv)
6
7
2S=2+2
7+2
72+2
73+ .......
=21
11
7
=27
6
4S =87
6
3=7
3
3
( )
[ ] ( )
( ) ( )
a1,a2,a3...areinA.P.(Letcommondifferenceisd1)
b1,b2,b3...areinA.P.(Letcommondifferenceisd2)
anda1=2,a10 =3,a1b1=1=a10b10
a1b1=1
b1=1
2
a10b10 =1
b10 =1
3
Now,a10 =a1+9d 1d1=1
9
b10 =b1+9d 2d2=1
9
1
31
2= 1
54
Now,a4=2+3
9=7
3
b4=1
23
54 =4
9
a4b4=28
27
[ ]
Question75
LetA1,A2,A3, ......beanincreasinggeometricprogressionofpositivereal
numbers.IfA1A3A5A7=1
1256andA2+A4=7
36,thenthevalueofA6+A8+A10is
equalto
[28-Jun-2022-Shift-1]
Options:
A.33
B.37
C.43
D.47
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question76
Ifnarithmeticmeansareinsertedbetweenaand100suchthattheratioof
thefirstmeantothelastmeanis1 :7anda+n =33,thenthevalueofnis:
[28-Jun-2022-Shift-2]
Options:
A.21
B.22
C.23
D.24
Answer:C
Solution:
Solution:
A1A3A5A7=1
1296
(A4)4=1
1296
A4=1
6
A2+A4=7
36
A2=1
36
A6=1
A8=6
A10 =36
A6+A8+A10 =43
a,A1,A2.......... . An,100
LetdbethecommondifferenceofaboveA.P.then
a+d
100 d=1
7
7a +8d =100....... (i)
anda+n=33
-------------------------------------------------------------------------------------------------
Question77
Letforn =1,2, ......, 50,Snbethesumoftheinfinitegeometricprogression
whosefirsttermisn2andwhosecommonratiois 1
(n+1)2.Thenthevalueof
1
26 +50
n=1Sn+2
n+1n1 isequalto____
[28-Jun-2022-Shift-2]
Answer:41651
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question78
Let{an}n=0
beasequencesuchthata0=a1=0andan+2=2an+1an+1
foralln 0.Then,
n=2
an
7nisequalto:
[29-Jun-2022-Shift-1]
Options:
A. 6
343
B. 7
216
C. 8
343
D. 49
216
Answer:B
Solution:
( )
and100 =a+ (n+1)d
100 =a+ (34 a)(100 7a)
8
800 =8a +7a2338a +3400
7a2330a +2600 =0
a=10,260
7,but a 260
7
n=23
Sn=n2
11
(n+1)2
=n(n+1)2
n+2= (n2+1) 2
n+2
Now 1
26 +
50
n=1
Sn+2
n+1n1
=1
26 +
50
n=1
(n2n) + 21
n+11
n+2
=1
26 +50 ×51 ×101
650 ×51
2+21
21
52
=1+25 ×17(101 3)
=41651
( )
{ ( ) }
( )
Solution:
-------------------------------------------------------------------------------------------------
Question79
Thesumoftheinfiniteseries1 +5
6+12
62+22
63+35
64+51
65+70
66+ ......isequalto:
[29-Jun-2022-Shift-2]
Options:
A. 425
216
B. 429
216
C. 288
125
D. 280
125
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question80
Let3,6,9,12,upto78termsand5,9,13,17,upto59termsbetwoseries.
Then,thesumofthetermscommontoboththeseriesisequalto_____
[29-Jun-2022-Shift-2]
Answer:2223
an+2=2an+1an+1&a0=a1=0
a2=2a1a0+1=1
a3=2a2a1+1=3
a4=2a3a2+1=6
a5=2a4a3+1=10
n=2
an
7n=a2
72+a3
73+a4
74+ ...
s=1
72+3
73+6
74+10
75+ ...
1
7s=1
73+3
74+6
75+ ...
6s
7=1
72+2
73+3
74+ ...
6s
49 =1
73+2
74+ ...
36s
49 =1
72+1
73+1
74+ ...
36s
49 =
1
72
11
7
36s
49 =7
49 ×6
s=7
216
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question81
Leta1=b1=1,an=an1+2andbn=an+bn1foreverynaturalnumber
nsl ant2.Then 15
n=1anbnisequalto
[25-Jul-2022-Shift-1]
Answer:27560
Solution:
Solution:
1stAP:
3,6,9,12, .......upto78terms
t78 =3+ (78 1)3
=3+77 ×3
=234
2ndAP:
5,9,13,17, ......upto59terms
t59 =5+ (59 1)4
=5+58 ×4
=237
Commonterm'sAP:
Firstterm = 9
CommondifferenceoffirstAP =3
AndcommondifferenceofsecondAP =4
∴Commondifferenceofcommonterms
AP =LCM(3,4) = 12
NewAP =9,21,33, ......
tn=9+ (n1)12 234
n237
12
n=19
S19 =19
2[2.9 + (19 1)12]
=19(9+108)
=2223
Given,
an=an1+2
anan1=2
∴Inthisseriesbetweenanytwoconsecutivestermsdifferenceis2.SothisisanA.P.withcommondifference2.
Alsogivena1=1
∴Seriesis = 1,3,5,7......
an=1+ (n1)2=2n 1
Alsobn=an+bn1
Whenn=2then
b2b1=a2=3
. b21=3 [Given b1=1]
b2=4
Whenn=3then
b3b2=a3
b34=5
b3=9
∴Seriesis = 1,4,9......
=12,22,32..... . n2
bn=n2
Now,
15
n=1
(an.bn)
=
15
n=1
[(2n 1)n2]
-------------------------------------------------------------------------------------------------
Question82
Thesum 21
n=1
3
(4n 1)(4n +3)isequalto
[25-Jul-2022-Shift-2]
Options:
A. 7
87
B. 7
29
C. 14
87
D. 21
29
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question83
ConsidertwoG.Ps.2,22,23, .....and4,42,43, ....of60andnterms
respectively.Ifthegeometricmeanofallthe60 +ntermsis(2)
225
8,then
k=1
nk(nk)isequalto:
[26-Jul-2022-Shift-1]
Options:
A.560
B.1540
C.1330
D.2600
Answer:C
Solution:
Solution:
=
15
n=1
2n3
15
n=1
n2
=2(13+23+ .. . 153) (12+22+ .. . 152)
=2×15 ×16
2
215(16) × 31
6
=27560
( ) ( )
21
n=1
3
(4n 1)(4n +3)=3
4
21
n=1
1
4n 11
4n +3
=3
4
1
31
7+1
71
11 + ... + 1
83 1
87
=3
4
1
31
87 =3
4
84
3.87 =7
29
[ ( ) ( ) ( ) ]
[ ]
GivenG.P's2,22,23, .. . .60terms
-------------------------------------------------------------------------------------------------
Question84
Theseriesofpositivemultiplesof3isdividedintosets:
{3}, {6,9,12}, {15,18,21,24,27}, ...Thenthesumoftheelementsinthe
11 th setisequalto__________.
[26-Jul-2022-Shift-1]
Answer:6993
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question85
If 10
k=1
k
k4+k2+1=m
n,wheremandnareco-prime,thenm +nisequalto________.
[26-Jul-2022-Shift-2]
Answer:166
Solution:
4,42, ... . nterms
Now,G.M = 2
225
8
(2.22...4.42...)
1
60 +n=2
225
8
2
n2+n+1830
6+n=2
225
8
n2+n+1830
60 +n=225
8
8n2217n +1140 =0
n=57
8,20,so n =20
20
k=1
k(20 k) = 20×20 ×21
220 ×21 ×41
6
=20 ×21
220 41
3=1330
( )
[ ]
{3×1}
1term
, {3×2,3×3,3×4}
3 terms
, {3×5,3×6,3×7,3×8,3×9}
5terms
, ...
11 th setwillhave1+ (10)2=21term
Alsoupto10 th settotal3×ktypetermswillbe1+3+5+ ...... + 19 =100−term
∴Set11 = {3×101,3×102, ..... . 3×121}
∴Sumofelements = 3× (101 +102 + ... + 121)
=3×222 ×21
2=6993
10
k=1
k
k4+k2+1
=1
2
10
k=1
1
k2k+11
k2+k+1.
[( )
-------------------------------------------------------------------------------------------------
Question86
DifferentA.P.'sareconstructedwiththefirstterm100,thelastterm199,
andintegralcommondifferences.Thesumofthecommondifferencesofall
suchA.P.'shavingatleast3termsandatmost33termsis________.
[26-Jul-2022-Shift-2]
Answer:53
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question87
Supposea1,a2, ..., an, ..beanarithmeticprogressionofnaturalnumbers.If
theratioofthesumoffirstfivetermstothesumoffirstninetermsofthe
progressionis5 :17and,110 <a15 <120,thenthesumofthefirstten
termsoftheprogressionisequalto
[27-Jul-2022-Shift-1]
Options:
A.290
B.380
C.460
D.510
Answer:B
Solution:
Solution:
=1
211
3+1
31
7+1
71
13 + ... + 1
91 1
111
=1
211
111 =110
2.111 =55
111 =m
n
m+n=55 +111 =166
[ ]
[ ]
d1=199 100
2I
d2=199 100
3=33
d3=199 100
4I
dn=199 100
i+1I
di=33 +11,9
SumofCD's = 33 +11 +9
=53
a1,a2, ... . anbeanA.Pofnaturalnumbersand
S5
S9
=5
17
5
2[2a1+4d ]
9
2[2a1+8d ]
=5
17
-------------------------------------------------------------------------------------------------
Question88
Letf (x) = 2x2x1andS = {n : | f(n) | 800}.Then,thevalueof
nSf(n)
isequalto_______.
[27-Jul-2022-Shift-1]
Answer:10620
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question89
LetthesumofaninfiniteG.P.,whosefirsttermisaandthecommonratiois
r,be5.Letthesumofitsfirstfivetermsbe 98
25.Thenthesumofthefirst21
termsofanAP,whosefirsttermis10ar,nth termisanandthecommon
differenceis10ar2,isequalto:
[27-Jul-2022-Shift-2]
Options:
A.21a11
B.22a11
C.15a16
D.14a16
Answer:A
Solution:
Solution:
34a1+68d =18a1+72d
16a1=4d
d=4a1
And110 <a15 <120
110 <a1+14d <120 110 <57a1<120
a1=2(∵aiN)
d=8
S10 =5[4+9×8] = 380
| f(n) | 800
⇒−800 2n2n1800
2n2n801 0
n6409 +1
4,6409 +1
4andnz
n= 19, 18, 17, ........., 19,20.
(2x2x1) = 2x2 x 1.
=2.2 (12+22+ ... + 192) + 2.20220 40
=10620
[ ]
LetfirsttermofG.P.beaandcommonratioisrThen, a
1r=5......(i)
-------------------------------------------------------------------------------------------------
Question90
2313
1×7+4333+2313
2×11 +6353+4333+2313
3×15 + + 303293+283273+ ... + 2313
15 ×63 isequalto
________.
[27-Jul-2022-Shift-2]
Answer:120
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question91
Considerthesequencea1,a2,a3, ...suchthata11,a22and
an+2=2
an+1
+anforn 1,2,3, ...If
a1+1
a2
a3
a2+1
a3
a4
a3+1
a4
a5.. .
a30 +1
a31
a32 =2α(61C31),thenαisequal
to:
[28-Jul-2022-Shift-1]
Options:
A.30
B.31
C.60
( ) ( ) ( ) ( )
a(r51)
(r1)=98
25 1r5=98
125
r5=27
125,r=3
5
3
5
Then, S21 =21
2[2×10ar +20 ×10ar2]
=21[10ar +10.10ar2]
=21a11
( )
Tn=
n
k=1
[(2k)3 (2k 1)3]
n(4n +3)
=
n
k=1
4k2+ (2k 1)2+2k(2k 1)
n(4n +3)
=
n
k=1
(12k26k +1)
n(4n +3)
=2n(2n2+3n +1) 3n23n +n
n(4n +3)
=n2(4n +3)
n(4n +3)=n
Tn=n
Sn=
15
n=1
Tn=15 ×16
2=120
D.61
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question92
Forp,qRs considertherealvaluedfunctionf (x) = (xp)2q,xRand
q>0.Leta1,a2a3anda4beinanarithmeticprogressionwithmeanpand
positivecommondifference.Iff (ai) | = 500foralli =1,2,3,4,thenthe
absolutedifferencebetweentherootsoff (x) = 0is_______.
[28-Jul-2022-Shift-1]
Answer:50
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question93
Letx1,x2,x3, ..., x20beingeometricprogressionwithx1=3andthecommon
an+2=2
an+1
+an
anan+1+1=an+1an+21
an+2an+1anan+1=2
For
n=1 a3a2a1a2=2
n=2 a4a3a3a2=2
n=3 a5a4a4a3=2
.
.
.
n=nan+2an+1anan+1=2
an+2an+1=2n +a1a2
Now,
(a1a2+1)
a2a3
(a2a3+1)
a3a4
(a3a4+1)
a4a5
......⋅ (a30a31 +1)
a31a32
=3
45
67
8......⋅ 61
62
=260(61C31)
a1,a2,a3,a4
a2=p3d ,a2=pd,a3=p+danda4=p+3d
Whered>0
| f(ai) | = 500
| 9d 2q| = 500
and|d2q| = 500
either9d 2q=d2q
d=0notacceptable
9d 2q=qd2
5d 2q=0
Rootsoff(x) = 0arep+ qandp q
∴absolutedifferencebetweenroots= | 2q| = 50
ratio 1
2.Anewdataisconstructedreplacingeachxiby(xii)2.Ifxisthe
meanofnewdata,thenthegreatestintegerlessthanorequaltoxis________.
[28-Jul-2022-Shift-1]
Answer:142
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question94
6
312 +10
311 +20
310 +40
39+ ... + 10240
3=2nm,wheremisodd,thenm .nisequal
to_______.
[28-Jul-2022-Shift-2]
Answer:12
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question95
x1,x2,x3, ...., x20areinG.P.
x1=3,r=1
2
x=xi
22xii+i2
20
=1
20 12 1 1
240 6 4 11
218 +70 ×41
S=1+21
2+31
22+ ...
S
2=1
2+2
22+ ....
.S
2=2 1 1
220 20
220 =411
218
[x] = 2858
20 12
240 66
218 1
20
=142
[ ( ) ( ) ]
{
( ) }
[ ( ) ]
1
312 +520
312 +21
311 +22
310 + ..... + 211
3=2nm
1
312 +51
312
((6)21)
(61)=2nm
1
312 +5
5
1
312 212 312 1
312 =2nm
1
312 +212 1
312 =2nm
2nm=212
m=1andn=12
mn=12
( )
( )
( )
If 1
(20 a)(40 a)+1
(40 a)(60 a)+ ... + 1
(180 a)(200 a)=1
256,thenthemaximumvalueof
ais:
[29-Jul-2022-Shift-1]
Options:
A.198
B.202
C.212
D.218
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question96
Leta1,a2,a3, ...beanA.P.If
r=1
ar
2r=4,then4a2isequalto________.
[29-Jul-2022-Shift-1]
Answer:16
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question97
If 1
2×3×4+1
3×4×5+1
4×5×6+ ... + 1
100 ×101 ×102 =k
101,then34kisequalto________.
[29-Jul-2022-Shift-1]
1
20
1
20 a1
40 a+1
40 a1
60 a+ ... + 1
180 a1
200 a=1
256
1
20
1
20 a1
200 a=1
256
1
20
180
(20 a)(200 a)=1
256
(20 a)(200 a) = 9.256
ORa2220a +1696 =0
a=212,8
( )
( )
( )
Given
S=a1
2+a2
22+a3
23+a4
24+ ..... .
1
2S=a1
22+a2
23+ ........ .
S
2=a1
2+(a2+a1)
22+(a3+a2)
23+ ..... .
S
2=a1
2+d
2
a1+d=a2=44a2=16
Answer:286
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question98
Let{an}n=0
beasequencesuchthata0a10and
an+2=3an+12an+1, n0.
Thena25a23 2a25a22 2a23a24 +4a22a24isequalto
[29-Jul-2022-Shift-2]
Options:
A.483
B.528
C.575
D.624
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question99
20
r=1(r2+1)(r!)isequalto
[29-Jul-2022-Shift-2]
Options:
S=1
2×3×4+1
3×4×5+1
4×5×6+ ... + 1
100 ×101 ×102
=1
(31) 1
1
2×31
101 ×102
=1
2
1
61
101 ×102
=143
102 ×101 =k
101
34k =286
[ ]
( )
a0=0,a1=0
an+2=3an+12an+1:n0
an+2an+1=2(an+1an) + 1
n=0 a2a1=2(a1a0) + 1
n=1 a3a2=2(a2a1) + 1
n=2 a4a3=2(a3a2) + 1
n=n an+2an+1=2(an+1an) + 1
(an+2a1) 2(an+1a0) (n+1) = 0
an+2=2an+1+ (n+1)
nn2
an2an1=n1
Now a25a23 2a25a22 2a23a24 +4a22a24
= (a25 2a24)(a23 2a22) = (24)(22) = 528
A.22! 21!
B.22! 2(21!)
C.21! 2(20!)
D.21! 20!
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question100
Ife(cos2x+cos4x+cos6x+ .. . )loge2satisfiestheequationt29t +8=0,
thenthevalueof 2 sin x
sin x + 3 cos x 0<x<π
2is
[24-Feb-2021Shift1]
Options:
A.23
B. 3
2
C.3
D. 1
2
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
( )
Given,
20
r=1
(r2+1)(r!)
Let,f(r) = (r2+1)(r!)
= (r2)(r!) + r!
=r(rr!) + r!
=r[(r+11)r!] + r!
=r[(r+1)r! r!] + r!
=r[(r+1)! (r!)] + r!
=r(r+1)! r(r!) + r! = (r+22)(r+1)! r(r!) + r!
= (r+2)(r+1)! 2(r+1)! [(r+11)(r!)] + r!
= (r+2)! 2(r+1)! (r+1)! + r! + r!
= (r+2)! 3(r+1)! + 2r!
= [(r+2)! (r+1)!] 2[(r+1)! r!]
20
r=1
f(r)
=
20
r=1
[(r+2)! (r+1)!] 2
20
r=1
[(r+1)! r!]
= [ (22! + 21! + 20! + .... + 4! + 3!) (21! + 20! + 19! + .... + 3! + 2!] 2[(21! + 20! + ... + 3! + 2!) (20! + 19! + ..... . 1!)]
= [(22!) (2!)] 2[(21)! (1!)]
=22! 2! 2 (21)! + 2.1!
=22! 2 (21)!
e(cos2x+cos4x+ .. . )ℓn2 =2cos2x+cos4x+ .. . =2cot2x
Nowt29t +9=0t=1,8
2cot2x=1,8cot2x=0,3
2 sin x
sin x + 3 cos x =2
1+ 3 cot x =2
4=1
2
Question101
LetA = { x:xis3-digitnumber]B = {x:x=9k +2,kI}and
C= {x:x=9k + ℓ, kI, I,O<ℓ<9}forsomeℓ(0<ℓ<9)Ifthesumof
alltheelementsofthesetA (BC)is274 ×400,thenisequalto
[24-Feb-2021Shift1]
Answer:5
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question102
Thesumoffirstfourtermsofageometricprogression(G.P.)is 65
12andthe
sumoftheirrespectivereciprocalsis 65
18.Iftheproductoffirstthreetermsof
theG.P.is1andthethirdtermisα1then2αis
[2021,24Feb.Shift-II]
Answer:9
Solution:
BandCwillcontainthreedigitnumbersoftheform9k +2and9k + ℓrespectively.Weneedtofindsumofallelementsintheset
B∪Ceffectively.
Now,S(BC) = S(B) + S(C) S(BC)whereS(k)denotessumofelementsofsetk.
Also B= {101,110, ..... . 992}
S(B) = 100
2(101 +992) = 54650
Case-I:Ifℓ = 2
then B C=B
S(BC) = S(B)
whichisnotpossibleasgivensumis
274×400=109600
Case-II:Ifℓ 2
then B C=φ
S(BC) = S(B) + S(C) = 400 ×274
54650 +
110
k=11
9k + = 109600
9
110
k=11
k+
110
k=11
= 54950
9100
2(11 +110) + ℓ(100) = 54950
54450 +100 = 54950
= 5
( )
LetfournumbersinGPbea,ar,ar2,ar3.
Accordingtothequestion,
a+ar +ar2+ar3=65
12 ⋅⋅⋅⋅⋅⋅ (i)
and 1
a+1
ar +1
ar2+1
ar3=65
18
-------------------------------------------------------------------------------------------------
Question103
Thesumoftheseries
n=1
n2+6n +10
(2n +1)! isequalto
[2021,26Feb.Shift-II]
Options:
A. 41
8e+19
8e110
B. 41
8e19
8e110
C. 41
8e+19
8e1+10
D.41
8e+19
8e110
Answer:B
Solution:
Solution:
1
a
1+r+r2+r3
r3=65
18 ⋅⋅⋅⋅⋅⋅ (ii)
DividingEq.(i)by(ii),weget
a(1+r+r2+r3)
1
a
(1+r+r2+r3)
r3
=65 12
65 18
a2r3=18
12
a2r3=3
2
Also,productoffirstthreeterms = 1
a×ar ×ar2=1
a3r3=1
a3×3
2a2=1r3=32
a2
a=2
3
andr3=32
(23)2=3
2
3⇒ r=3
2
Accordingtothequestion,thirdterm
α=ar2=2
3×3
2×3
2=3
2∴ =2×3
2=3thirdterm
( )
[ ]
( )
Let
n=1
n2+6n +10
(2n +1)! =S
=
n=1
4n2+24n +40
4(2n +1)!
=
n=1
(2n +1)2+ (2n +1) 10 +29
4(2n +1)!
=1
4
n=1
(2n +1)2
(2n +1)(2n)! +
n=1
(2n +1) 10
(2n +1)(2n)!+
n=1
29
(2n +1)!
=1
4
n=1
(2n +1)
(2n)! +
n=1
10
(2n)! +
n=1
29
(2n +1)! ⋅⋅⋅⋅⋅⋅ (1)
Now,
=
n=1
(2n +1)
(2n)! =
n=1
2n
(2n)! +
n=1
1
(2n)!
=
n=1
1
(2n 1)! +
n=1
1
(2n)!
Now,
[ ]
[ ]
-------------------------------------------------------------------------------------------------
Question104
Thesumoftheinfiniteseries1 +2
3+7
32+12
33+17
34+22
35+ ...isequalto
[2021,26Feb.Shift-1]
Options:
A. 13
4
B. 9
4
C. 15
4
D. 11
4
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
=
n=1
1
(2n 1)! =1
1!+1
3!+1
5!+ ... =
e1
e
2⋅⋅⋅⋅⋅⋅ (ii)
and
n=1
1
(2n)! =1
2!+1
4!+1
6!+ ... =
e+1
e2
2⋅⋅⋅⋅⋅⋅ (iii)
and
n=1
1
(2n +1)! =1
3!+1
5!+1
7!+ ... =
e1
e2
2⋅⋅⋅⋅⋅⋅ (iv)
UsingEqs.(ii),(iii),(iv)in(i),
S=1
4
e1
e
2+11&
e+1
e2
2
+29&
e1
e2
2
=1
4
e
21
2e +11e
2+11
2e +29e
229
2e 4
=41e
819
8e 10
[ ( )
( ) ]
[ ]
Given,S=1+2
3+7
32+12
33+ ...
Let,S1=2
3+7
32+12
33+ ...
Multiply13inseriesEq.(i),
S1
3=2
32+7
33+12
34+ ...
SubtractEq.(ii)fromEq.(i),weget
S1S1
3=2
3+5
32+5
33+ ...
2S1
3=2
3+5
32+5
33+ ...
=2
3+532
1135
32+5
33+ ...isageometricserieswithr=13,sumuptoinfinityofthisseriesis a
1r,wherea= first
term]
=2
3+5
6=9
6=3
2
S1=3
2×3
2=9
4
S=1+S1
=1+9
4=13
4
[ ]
[ ] [
[ ]
Question105
Ifthearithmeticmeanandgeometricmeanofthepthandqthtermsofthe
sequence16,8, 4,2,...satisfytheequation
4x29x +5=0,thenp +qisequalto
[2021,26Feb.Shift-II]
Answer:10
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question106
Inanincreasinggeometricseries,thesumofthesecondandthesixthterm
is 25
2andtheproductofthethirdandfifthtermis25.Then,thesumof4th,
6thand8thtermsisequalto
[2021,26Feb.Shift-1]
Options:
A.30
B.26
IfAM andGM satisfytheequation4x29x +5=0,thenAMandGMarenothingbutrootsofthisquadraticequation,
4x29x +5=0
4x24x 5x +5=0
4x(x1) 5(x1) = 0
(x1)(4x 5) = 0
x=1,5
4
Then,AM =5
4and GM =1[∵AM GM ]
Again,thegivenseriesis
16,8, 4,2.....
whichisageometricprogressionserieswithcommonratio 1
2,then
pthterm = 16 1
2
p1=tp
qthterm = 16 1
2
q1=tq
Arithmeticmean = 5
4
tp+tq
2=5
4
Geometricmean = 1
tptq=1
tptq=1
⇒(16)1
2
p1(−16)1
2
q1=1
(−16)21
2
p+q2=1
(−24)21
2
p+q2=1
(−2)8(+1)p+q2
(−2)p+q2=1
⇒(2)8(+1)p+q2= (−2)p+q2
(−2)8= (−2)p+q2
p+q2=8
p+q=10
( )
( )
( ) ( )
( )
( )
C.35
D.32
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question107
Theminimumvalueoff (x) = aax+a1ax,wherea,xRanda >0,isequalto
[2021,25Feb.Shift-II]
Options:
A.a +1
B.a +1
a
C.2a
D.2a
Answer:C
Solution:
Solution:
Letthefirsttermofgeometricseriesbe'aandcommonratiobe'r'.
Then,nthtermofgivenseriesisgivenas
Tn=arn1
Now,giventhatsumofsecondandsixthtermis25/2.
i.e.T2+T6=25 2
ar +ar5=25 2
ar(1+r4) = 25 2⋅⋅⋅⋅⋅⋅ (i)
Also,giventhatproductofthirdandfifthtermis25.
l rl i.e. (T3)(T5) = 25
(ar2)(ar4) = 25
a2r6=25 ⋅⋅⋅⋅⋅⋅ (ii)
SquaringEq.(i),wegeta2r2(1+r4)2=25
2
2⋅⋅⋅⋅⋅⋅ (iii)
DivideEq.(ii)by(iii),
a2r2(1+r4)2
a2r6=(25)2
4(25)
(1+r4)2
r4=25
4
4(1+r4)2=25r4
4(1+r8+2r4) = 25r4
4r817r4+4=0
4r816r4r4+4=0
4r4(r44) 1(r4+ (−4)) = 0
(r44)(4r41) = 0
Gives, r4=40rr4=14
Wehavetofindsumof4th,6thand8thterm,i.e.
T4+T6+T8=ar3+ar5+ar7
=ar(r2+r4+r6)
=ar3(1+r2+r4) ⋅⋅⋅⋅⋅⋅ (iv)
UsingEq.(ii),
(ar3)2=25
ar3=5
Also,wetaker4=4becausegivenseriesisincreasingandr2=2.
T4+T6+T8=5(1+2+4)
=5(7) = 35
( )
Wealreadyknow,Arithmeticmean≥Geometricmean,
-------------------------------------------------------------------------------------------------
Question108
If0 <θ,φ<π
2,x=
n=0cos2nθ1y =
n=0sin2nφand
z=
n=0cos2nθsin2nφ,then
[2021,25Feb.Shift-1]
Options:
A.xy z= (x+y)z
B.xy +yz +zx =z
C.xyz =4
D.xy +z= (x+y)z
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
LetustakeAMandGMoftwotermsaa*anda1ax,
AM =aax+a1ax
2
and GM =aaxa1ax
AM GM aax+a1ax
2aaxa1ax
aax+a1ax2a1
Minimumvalueof f (x) = aax+a1axis
2a.
Given,x=
n=0
cos2nθ
y=
n=0
sin2nφ
z=
n=0
cos2nθsin2nφ
x=1+cos2θ+cos4θ+ .. .
x=1
1cos2θ=cosec2θ
y=1+sin2φ+sin4φ+ .. .
x=1
1cos2θ=cosec2θ⋅⋅⋅⋅⋅⋅ (i)
y=1+sin2φ+sin4φ+ .. .
y=1
1sin2φ = sec2φ⋅⋅⋅⋅⋅⋅ (ii)
z=1+cos2θsin2φ+cos4θsin4φ + .. .
z=1
1cos2θsin2φ...(iii)
FromEqs.(i),(ii)and(iii),weget
z=1
111
x11
y
cos2θ=11
x
sin2φ=11
y
z=xy
xy (x1)(y1)
z=xy
xy xy +x+y1
xz +yz z=xy
xy +z= (x+y)z
( ) ( ) [ ]
Question109
LetA1,A2,A3, ........besquares,suchthatforeachn 1,thelengthofthe
sideofAnequalsthelengthofdiagonalofAn+1.IfthelengthofA1is12cm,
thenthesmallestvalueofnforwhichareaofAnislessthanone,is..........
[2021,25Feb.Shift-I]
Answer:9
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question110
Consideranarithmeticseriesandageometricserieshavingfourinitialterms
fromtheset{11,8,21,16,26,32,4}.Ifthelasttermsoftheseseriesare
themaximumpossiblefourdigitnumbers,thenthenumberofcommon
termsinthesetwoseriesisequalto.........
[2021,16MarchShift-1]
Accordingtothequestion,lengthofsideofA1squareis12 mathrm ~cm.
∵SidelengthsareinGP.
Tn=12
(√2)n1
(Sideofnthsquarei.e.An)
Area = ( Side )2=12
(√2)n1
2=144
2n1
Accordingtothequestion,theareaofAnsquare<1
2n1>144
Here,thesmallestpossiblevalueofis = 9.
( )
Answer:3
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question111
Let 1
16,aandbbeinG.P.and 1
a,1
b,6bein(a)P.,wherea,b>0.Then,
72(a+b)isequalto.........
[2021,16MarchShift-II]
Answer:14
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question112
Ifα,βarenaturalnumbers,suchthat100α199β = (100)(100) + (99)(101)+
(98)(102) + ... + (1)(199),thentheslopeofthelinepassingthrough(α,β)
andoriginis
[2021,18MarchShift-1]
Options:
Given,set{11,8,21,16,26,32,4}
Byobservation,wecansaythat
AP = {11,16,21,26, ...}
GP = {4,8,16,32, ...}
5m +6=4.2n1
5m +6=2n+1
So,(2n+16)shouldbeamultipleof5.Theunitdigitof2kis2,4,6,8.So,when6issubtractedfrom2n+1,thepossibleunit
digitswillbe6,8,0,2.Only0isdivisibleby5.Hence,2n+1unitdigithastobe6.
2n+1=24,28,212,216...
As,216willnotbea4digitnumber,so,commonterms = {16,256,4096}
∴Numberofcommonterms = 3
Given, GP =1
16,a,b
a2=b
16
andgiven,AP =1a,1b,6
2
b=1
a+6
2
16a2=1
a+6
1
8a2=1+6a
a
1=8a(1+6a)
48a2+8a 1=0
(4a +1)(12a 1) = 0
a= 14 or 1 12
Asperthequestion,a>0
a=112
b=16a2=161
144 =1
9
72(a+b) = 72 1
12 +1
9
=6+8
=14
( )
A.540
B.550
C.530
D.510
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question113
1
321+1
521+1
721+ ... + 1
(201)21
isequalto
[2021,18MarchShift-1]
Options:
A. 101
404
B. 25
101
C. 101
408
D. 99
400
Answer:B
Solution:
Solution:
Given,
100α199 .β= (100)(100) + (99)(101)
+(98)(102) + ... + (1)(199)
100α199β =
99
x=0
(100 x)(100 +x)
=
99
x=0
(1002x2)
=
99
x=0
(100)2
99
x=0
(x)2
= (100)399 ×100 ×199
6
(100)α (199)β= (100)3 (199)(1650)
Oncomparing,wegetα=3,β=1650
Then,theslopeofthelinepassing
through(α,β)andoriginis
=β0
α0=β
α=1650
3=550
1
321+1
521+1
721+ ... + 1
(201)21
=
100
r=1
1
(2r +1)21
=
100
r=1
1
4r2+4r +11
=
100
r=1
1
2r(2r +2)=
100
r=1
1
4(r)(r+1)
-------------------------------------------------------------------------------------------------
Question114
LetS1bethesumoffirst2ntermsofanarithmeticprogression.LetS2bethe
sumoffirst4ntermsofthesamearithmeticprogression.If(S2S1)is1000,
thenthesumofthefirst6ntermsofthearithmeticprogressionisequalto
[2021,18MarchShift-11]
Options:
A.1000
B.7000
C.5000
D.3000
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question115
Iflog32,log3(2x5), log32x7
2areinanarithmeticprogression,thenthe
valueofxisequalto
[2021,27JulyShift-1]
Answer:3
Solution:
( )
=1
4
100
r=1
1
r1
r+1
=1
4
1
11
2+1
21
3+1
31
4+... + 1
100 1
101
=1
411
101 =1
4×100
101
=25
101
( )
[ ( ) ( ) ( ) ( ) ]
[ ]
Given,S1=S2nandS2=S4n
andS2S1=1000 S4n S2n =1000
4n
2[2a + (4n 1)d] 2n
2[2a + (2n 1)d] = 1000
2n[2a + (4n 1)d] n[2a + (2n 1)d] = 1000
2an +n(8n 22n +1)d=1000
2an +n(6n 1)d=1000
n[2a + (6n 1)d] = 1000
S6n =6n
2[2a + (6n 1)d]
=6
2× (1000)
=6×500 =3000
Question116
Iftan π
9,x,tan
18 areinarithmeticprogressionandtan π
9,y,tan
18
arealsoinarithmeticprogression,then|x2y|isequalto
[2021,27JulyShift-II]
Options:
A.4
B.3
C.0
D.1
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question117
LetSnbethesumofthefirstntermsofanarithmeticprogression,If
S3n =3S2n,thenthevalueof S4n
S2n
is
[2021,25JulyShift-1]
( ) ( ) ( ) ( )
log32,log3(2x5), log32x7
2AP
2log3(2x5) = log32+log32x7
2
log3(2x5)2=log322x7
2
(2x5)2=22x7
(2x)2+25 10 2x22x+7=0
(2x)212 2x+32 =0
(2x4)(2x8) = 0
2x=4or8x=2or3
Ifx=2,thenlog3(2x5) = log3(225)
Here,argumentisnegative,so,x2.
Hence,x=3
( )
( )
[ ( ) ]
Iftan π
9,x,tan
18 areinAP.
So,x=1
2tan π
9+tan
18
( ∵ifa,b,careinAP,so,b=a+c
2
Andtan π
9,y1tan
18 areinAP.
Now,x2y =1
2tan π
9+tan
18 − tan π
9+tan
18
| x2y | =
cot π
9tan π
9
2tan
18
tan
18 =cot
9
and tan
18 =cot π
9
=cot
9cot
9=0
cot 2 A =2cot2A1
2 cot A
( ) ( )
[ ( ) ]
)
( ) ( )
[ ] ( )
| | { }
| |
[ ]
Options:
A.6
B.4
C.2
D.8
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question118
LetSndenotethesumoffirstntermsofanarithmeticprogression.If
S10 =530,S5=140,thenS20 S6isequalto
[2021,22JulyShift-II]
Options:
A.1862
B.1842
C.1852
D.1872
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question119
Thesumofalltheelementsintheset{n {1,2, .. . 100}:HCFofnand
2040is1}isequalto
[2021,22JulyShift-II]
LetSn=An2+Bn =n(An +B)S3n =3S2n
3n[A(3n) + B] = 32n [A(2n) + B]
3An +B=4An +2B
An +B=0
S4n
S2n
=4n[A(4n) + B]
2n[A(2n) + B]
=24An An
2An An
=2×3=6
( )
Sn=An2+Bn
S10 =100A +10B =530
S5=25A +5B =140Solvingbothequations,wegetB=3and
A=5
Sn=5n2+3n
S20 S6=5(20262) + 3(20 6)
=526.14 +314
=14(130 +3) = 14 ×133 =1862
Answer:1251
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question120
Ifsumofthefirst21termsoftheseries
log912x+log913x+log914x+ .....wherex >0
is504,thenxisequalto
[2021,20JulyShift-II]
Options:
A.243
B.9
C.7
D.81
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question121
Fork N ,let
1
α(α+1)(α+2)... . (α+20)=20
k=0
Ak
α+k
whereα >0.Thenthevalueof100 A14 +A15
A13
2
[2021,20JulyShift-II]
( )
n {1,2,3, ..... . 100}
2040 =23×3×5×17
IfHCFofnand2040is1,nshouldnotbeamultipleof2,3,5,17.
n { 1,7,11,13,19,23,29,31,37,41,43,47,53,59,61,67,71,73,77,79,83,89,91,97 }
n= | 1251|
LetS=log912x+log913x+ ...
Usingproperty,logabx=1
blogax
S=2log9x+3log9x+ ... + 22log9x
=log9x(2+3+4+ ... + 22)
=log9x21
2(4+20) = log9x(21 ×12)
S=252log9x
Given, S=504,then
252log9x=504
log9x=2
x= (9)2=81
[ ]
Answer:9
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question122
Ifthevalueof
1+2
3+6
32+10
33
+.... upto
log(0.25)
1
3+1
32+1
33+ ...... upto
isl ,thenl 2isequalto.......
[2021,25JulyShift-I]
Answer:3
Solution:
( ) ( )
Given,
1
α(α+1)... + (α+20)=
20
k=0
Ak
α+k,α>0
1
α(α+1).. . (α+20)=A0
α+A1
α+1
+... + A20
α+20
A14 =1
(−14)(−13).. . (−1)(1).. . (6)=1
14!6!
A15 =1
(−15)(−14).. . (−1)(1).. . (5)=1
15!5!
A13 =1
(−13).. . (−1)(1).. . (7)=1
13!7!
A14
A13
=13!7!
14!6!=7
14 =1
2
A15
A13
=13!7!
15!5!=42
15 ×14 =1
5
100 A14
A13
+A15
A13
2
=100 1
2+1
5
2
=100 3
10
2=9
( ) ( )
( )
-------------------------------------------------------------------------------------------------
Question123
Let{an}n=1
beasequencesuch
thata1=1,a2=1andan+2=2an+1+anforalln 1.Thenthevalueof
47
n=1
an
23nisequalto....
[2021,20JulyShift-II]
Answer:7
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question124
Leta1,a2,a3......bean(a)P.
L2=3
Let
n=1
an
23n =xi.e.
n=1
an
8n=x
Given,an+2=2an+1+an
Dividethewholeby8n,
an+2
8n=2an+1
8n+an
8n
82an+2
8n+2=82an+1
8n+1+an
8n
64 an+2
8n+2=16 an+1
8n+1+an
8n
Now,takethesummation,
64
n=1
an+2
8n+2=16
n=1
an+1
8n+1+
n=1
an
8n⋅⋅⋅⋅⋅⋅ (i)
n=1
an
8n=x
i.e. a1
8+a2
82+a3
83+a4
84+ ... = x
a3
83+a4
84+ ... = xa1
8a2
82
n=1
an+2
8n+2=xa1
8a2
82⋅⋅⋅⋅⋅⋅ (ii)
Again, a2
82+a3
83+ ... = xa1
8
n=1
an+1
8n+1=xa1
8⋅⋅⋅⋅⋅⋅ (iii)
FromEqs.(i),(ii)and(iii),
64 x a1
8a2
64 =16 x a1
8+x
Usea1=1=a2
64 x 1
81
64 =16 x 1
8+x
64x 9=2(8x 1) + x
64x 16x x=9247x =7
47 an
23n =7
( ) ( )
( ) ( )
( ) ( )
If a1+a2+ ... + a10
a1+a2+ ..... + ap
=100
p2,p10,then a11
a10
isequalto
[2021,31Auq.Shift-II]
Options:
A. 19
21
B. 100
121
C. 21
19
D. 121
100
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question125
Thenumberof4-digitnumberswhichareneithermultipleof7normultiple
of3is
[2021,31Aug.Shift-II]
Answer:5143
Solution:
a1+a2+ ... + a10
a1+a2+ ... + ap
=100
p2
S10
Sp
=100
p2Sp=S10 p2
100
a11
a10
=S11 S10
S10 S9
=
S10 121
100 S10
S10 S1081
100
=
121
100 1
181
100
=21
19
Total4-digitnumber
=9×10 ×10 ×10 =9000
4-digitnumberdivisibleby7
1001,1008, ..., 9996
Numberof4-digitnumberdivisibleby7
=9996 1001
7+1=1286
4-digitnumberdivisibleby3
1002,1005, ..., 9999
Numberof4-digitnumberdivisibleby3
=9999 1002
3+1=3000
4digitnumberdivisibleby21
1008,1031, ..., 9996
Numberof4-digitnumberdivisibleby21
=9996 1008
21 +1=429
∴Numberof4-digitnumbersneitherdivisibleby7nor3
=9000 1286 3000 +429 =5143
Question126
Threenumbersareinanincreasinggeometricprogressionwithcommon
ratior.Ifthemiddlenumberisdoubled,thenthenewnumbersareinan
arithmeticprogressionwithcommondifferenced .IfthefourthtermofGPis
3r2,thenr2d isequalto
[2021,31Aug.Shift-I]
Options:
A.7 73
B.7 + 3
C.7 3
D.7 +33
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question127
If0 <x<1,then
3
2x2+5
3x3+7
4x4+ ......,isequalto
[2021,27Aug.Shift-1]
Options:
A.x 1+x
1x+loge(1x)
B.x 1x
1+x+loge(1x)
C. 1x
1+x+loge(1x)
D. 1+x
1x+loge(1x)
( )
( )
Letthreenumbersbe a
r,a,ar.
Accordingtothequestion, a
r,2a,ar AP
4a =ar +a
rr+1
r=4
r24r +1=0r=2± 3
T4ofGP =3r2
3r2=ar2
a=3
r=2+ 3
d=2a a
r=33
r2d= (2+ 3)233
=7+4333=7+ 3
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question128
Ifx,yR,x>0
y=log10x+log10x13+log10x19+ ...
upto∞termsand
2+4+6+ ... + 2y
3+6+9+ ... + 3y =4
log10x,thentheorderedpair(x,y)isequalto
[2021,27Aug.Shift-1]
Options:
A.(106,6)
B.(104,6)
C.(102,3)
D.(106,9)
Answer:D
Solution:
Solution:
Wehave,
3
2x2+5
3x3+7
4x4+ ...
=21
2x2+21
3x3+21
4x4+ ...
=2(x2+x3+x4+ ...) x2
2+x3
3+x4
4+ ...
=2x2
1x [−loge(1x) x]
[usingsumofinfiniteGP =a
1randlogarithmicseries]
=2x2
1x+x+loge(1x)
=2x2+xx2
1x+loge(1x)
=x2+x
1x+loge(1x)
=x1+x
1x+loge(1x)
( ) ( ) ( )
( )
( )
Given, 2+4+6+ ... + 2y
3+6+9+ ... + 3y =4
log10x
2(1+2+3+ ... + y)
3(1+2+3+ ... + y)=4
log10x
2
3=4
log10xlog10x=6
x=106
Now,y=log10x+log10x
1
3+log10x
1
9+ ...uptoterms.
=log10 xx
1
3x
1
9... terms
=log10x
1+1
3+1
9+ .. . terms
( )
-------------------------------------------------------------------------------------------------
Question129
If0 <x<1andy =1
2x2+2
3x3+3
4x4+ ...thenthevalueofe1+yatx =1
2is
[2021,27Aug.Shift-II]
Options:
A. 1
2e2
B.2e
C. 1
2e
D.2e2
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question130
IfthesumofaninfiniteGPa,ar,ar2,ar3, ....is15andthesumofthe
squaresofitseachtermis150,thenthesumofar2,ar4,ar6, ...is
[2021,26Aug.Shift-1]
Options:
A.5 2
B.1 2
C.25 2
=log10x
1
11
3=log10x32
=log10(106)
3
2[∵x=106]
y=6×3
2=9
x=106,y=9
(x,y) = (106,9)
y=1
2x2+2
3x3+3
4x4+ ...
y=11
2x2+11
3x3+11
4x4+
= (x2+x3+x4+ ...) x2
2+x3
3+x4
4+ ...
=x2
1x+xx+x2
2+x3
3+x4
4...
y=x
1x+ln(1x)
Putx=1
2,weget
y=1ln 2
Then,e1+y=e1+1ln 2 =e2ln 2
=e2eln 21
=1
2e2
( ) ( ) ( )
( )
( )
D.9 2
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question131
Leta1,a2......, a10beanAPwithcommondifference3andb1,b2, ......, b10
beaGPwithcommonratio2.
Letck=ak+bk,k=1,2, ......, 10.Ifc2=12and
c3=13,then 10
k=1ckisequalto
[2021,26Aug.Shift-II]
Answer:2021
Solution:
Wehave,sumofinfiniteGPa,ar,ar2, ...is
S=a
1r=15 ⋅⋅⋅⋅⋅⋅ (i)
andsumofinfiniteGPa2,a2r2,a2r4, ...is
cS
=a2
1r2=150
a
1r
a
1+r=150 ⋅⋅⋅⋅⋅⋅ (ii)
DivideEq.(ii)byEq.(i)
a
1+r=10 ⋅⋅⋅⋅⋅⋅ (iii)
DivideEq.(iii)byEq.(i)
1r
1+r=10
15 =2
3
33r =2+2r
1=5r r=1
5
Now,puttingr=1
5inEq.(iii),weget
a
1+1
5
=10
5a
6=10 a=12
Now,sumofar2,ar4,ar6, ...,
S
′′ =ar2
1r2=
12 1
25
24
25
=1
2
( ) ( )
( )
a1,a2,a3, ..., a10areinAPcommon
difference = 3
b1,b2,b3, ..., b10areinGPcommonratio = 2
Since,c = ak+bk,k=1,2,3......, 10
c2=a2+b2=12
c3=a3+b3=13
Now,c3c2=1
(a3a2) + (b3b2) 1
3+ (2b2b2) 1
b2=4
a2=8
So,APis11,8,5, ....
-------------------------------------------------------------------------------------------------
Question132
Thesumof10termsoftheseries 3
12×22+5
22×32+7
32×42+ ...is
[2021,31Aug.Shift-1]
Options:
A.1
B.120/121
C.99/100
D.143 144
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question133
IfS =7
5+9
52+13
53+19
54+ ...,then160Sisequalto
[2021,31Aug.Shift-II]
Answer:305
Solution:
andGPis2,4,8, ....
Now,
10
k=1
ck=
10
k=1
ak+
10
k=1
bk
=10
2[22 +9(−3)] + 2210 1
21
=5(22 27) + 2(1023)
=2046 25 =2021
( ) ( )
3
12×22+5
22×32+7
32×42+ ...
=2212
12×22+3222
22×32+4232
32×42+ ....
=1
121
22+1
221
32+1
321
42+ .... 1
1021
112
=1
121
112=11
121 =120
121
( ) ( ) ( ) ( )
S=7
5+9
52+13
53+19
54+ ... + ⋅⋅⋅⋅⋅⋅ (i)
S
5=7
52+9
53+13
54+19
55+ .. . ⋅⋅⋅⋅⋅⋅ (ii)
SubtractingEq.(ii)fromEq.(i),
c4S
5=7
5+2
52+4
53+6
54+8
55+ .. .
4S
57
5=2
52+4
53+6
54+8
55+ .. . =K⋅⋅⋅⋅⋅⋅ (iii)
K
5=2
53+4
54+6
55+8
56+ .. . ... (iv)
-------------------------------------------------------------------------------------------------
Question134
Leta1,a2, ......, a21beanAPsuchthat 20
n=1
1
anan+1
=4
9.IfthesumofthisAPis
189,thena6a16isequalto
[2021,01Sep.Shift-II]
Options:
A.57
B.72
C.48
D.36
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question135
LetSn=1 (n1) + 2 (n2) + 3.
(n3) + ... + (n1) 1,n4 .
SubtractingEq.(iv)fromEq(iii),
4K
5=2
52+2
53+2
54+2
55+ .. .
4K
5=2
25
1
115=1
10 K=1
8
FromEq.(iii),
4S
57
5=1
8S=61
32
Now,106S =160×61
32 =305
( )
LetdbethecommondifferenceofanAP
a1,a2, ..., a21 and
20
n=1
1
anan+1
=4
9
20
n=1
1
an(an+d)=4
9
1
d
20
n=1
1
an
1
an+d=4
9
1
d
1
a1
1
a2
+1
a3
1
a4
+ ... + 1
a20
1
a21
=4
9
1
d
1
a1
1
a21
=4
9
1
d
a21 a1
a1a21
=4
9a1a21 =45
a1(a1+20d ) = 45 ⋅⋅⋅⋅⋅⋅ (i)
21
2(2a1+20d ) = 189
Alsosumoffirst21terms = 189
21
2(2a1+20d ) = 189
a1+10d =9⋅⋅⋅⋅⋅⋅ (ii)
ByEqs.(i)and(ii),wegeta1=3,d=35
ora1=15,d= 3
5
So, a6a16 = (a1+5d )(a1+15d ) = 72
( )
[ ]
( )
( )
Thesum
n=4
2Sn
n!1
(n2)! isequalto
[2021,01Sep.Shift-II]
Options:
A. e1
3
B. e2
6
C. e
3
D. e
6
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question136
ThenumberoftermscommontothetwoA.P.'s3,7,11, ...407and
2,9,16, ..., 709is______.
[NAJan.9,2020(II)]
Answer:14
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question137
Ifthe10 th termofanA.P.is 1
20andits20 th termis 1
10,thenthesumofits
( )
Sn=1 (n1) + 2(n2) + 3(n3)+... + (n1) 1,n4
=
n1
r=1
r(nr) = n(n21)
6
=n(n1)(n+1)
6
2Sn
n!=(n+1)
3(n2)!
n=4
2sn
n!1
(n2)!
=
n=4
n2
3(n2)! =1
3
n=4
1
(n3)!
=1
3
1
1!+1
2!+1
3!+ ... = e1
3
( )
( )
Firstcommontermofboththeseriesis23andcommondifferenceis7×4=28
∵Lastterm≤407⇒23 + (n1) × 28 407
(n1) × 28 384
n384
28 +1
n14.71
Hence,n=14
first200termsis:
[Jan.8,2020(II)]
Options:
A.50
B.50 1
4
C.100
D.100 1
2
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question138
Letf :RRbesuchthatforallx R, (21+x+21x), f(x)and(3x+3x)are
inA.P.,thentheminimumvalueoff (x)is:
[Jan.8,2020(I)]
Options:
A.2
B.3
C.0
D.4
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question139
FivenumbersareinA.P.,whosesumis25andproductis2520.Ifoneof
thesefivenumbersis1
2,thenthegreatestnumberamongstthemis:
T10 =1
20 =a+9d ...(i)
T20 =1
10 =a+19d ...(ii)
Solvingequations(i)and(ii),weget
a=1
200,d=1
200
S200 =200
2
2
200 +199
200 =201
2=100 1
2
[ ]
If21x+21+x,f(x), 3x+3xareinA.P.,then
f(x) = 21+x+21x+3x+3x
2
2f (x) = 2 2x+1
2x+3x+1
3x
UsingAM GM
f(x) 3
( )
( ) ( )
[Jan.7,2020(I)]
Options:
A.27
B.7
C. 21
2
D.16
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question140
Theproduct2
1
44
1
16 8
1
48 16
1
128 ...to∞isequalto:
[Jan.9,2020(I)]
Options:
A.2
1
2
B.2
1
4
C.1
D.2
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Let5termsofA.P.be
a2d ,ad,a,a+d,a+2d
Sum =25 5a =25 a=5
Product = 2520
(52d )(5d)5(5+d)(5+2d ) = 2520
(25 4d 2)(25 d2) = 504
625 100d 225d 2+4d 4=504
4d 4125d 2+625 504 =0
4d 4125d 2+121 =0
4d 4121d 24d 2+121 =0
(d21)(4d 2121) = 0
d= ±1,d= ± 11
2
d= ±1andd= 11
2,doesnotgive 1
2asaterm
d=11
2
∴Largestterm = 5+2d =5+11 =16
2
1
4+2
16 +3
48 + ... .
=2
1
4+1
8+1
16 + ... .
= 2
Question141
Letanbethen th termofaG.P.ofpositiveterms.
If 100
n=1a2n +1=200and 100
n=1a2n =100,then 200
n=1anisequalto:
[Jan.9,2020(II)]
Options:
A.300
B.225
C.175
D.150
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question142
Ifx =
n=0(−1)ntan2nθandy =
n=0cos2nθ,for0 <θ<π
4,then:
[Jan.9,2020(II)]
Options:
A.x(1+y) = 1
B.y(1x) = 1
C.y(1+x) = 1
D.x(1y) = 1
Answer:B
Solution:
Solution:
LetG.P.bea,ar,ar2......
100
n=1
a2n +1=a3+a5+ ..... + a201 =200
ar2(r200 1)
r21=200
100
n=1
a2n =a2+a4+ ..... + a200 =100
ar(r200 1)
r21=100
Fromequations(i)and(ii),r=2and
a2+a3+ ..... + a200 +a201 =300
r(a1+ ..... + a200) = 300
200
n=1
an=300
r=150
y=1+cos2θ+cos4θ+ ....
y=1
1cos2θ1
y=sin2θ
x=1tan2θ+tan4θ+ ....
x=1
1 (−tan2θ)=1
sec2θx=cos2θ
-------------------------------------------------------------------------------------------------
Question143
Thegreatestpositiveintegerk,forwhich49k+1isafactorofthesum
49125 +49124 + ... + 492+49 +1,is:
[Jan.7,2020(I)]
Options:
A.32
B.63
C.60
D.65
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question144
Leta1,a2,a3, ...beaG.P.suchthata1<0,a1+a2=4anda3+a4=16.If
9
i=1ai=,thenλisequalto:
[Jan.7,2020(II)]
Options:
A.-513
B.-171
C.171
D. 511
3
Answer:B
Solution:
Solution:
y=1
sin2θy=1
1x
y(1x) = 1
(49)126 1
48 =((49)63 +1)(4963 1)
48 Sn=a(rn1)
r1
K=63
[ ]
Since,a1+a2=4a1+a1r=4...(i)
a3+a4=16 a1r2+a1r3=16...(ii)
Fromeqn.(i),a1=4
1+randsubstitutingthevalueofa1,ineqn(ii),
4
1+r
r2
+4
1+r
r3
=16
4r2(1+r) = 16(1+r)
r2=4r= ±2
r=2,a1(1+2) = 4a1=4
3
r= 2,a1(12) = 4a1= 4
( ) ( )
-------------------------------------------------------------------------------------------------
Question145
Thecoefficientofx7intheexpression
(1+x)10 +x(1+x)9+x2(1+x)8+ ... + x10is:
[Jan.7,2020(II)]
Options:
A.210
B.330
C.120
D.420
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question146
Thesum, 7
n=1
n(n+1)(2n +1)
4isequalto_________.
[Jan.8,2020(II)]
Answer:504
Solution:
a
i=1
ai=a1(rq1)
r1=(−4)((−2)91)
21
=4
3(−513) = λ= 171
ThegivenseriesisinG.P.thenSn=a(1rn)
1r
(1+x)10 1x
1+x
11
1x
1+x
(1+x)10[(1+x)11 x11]
(1+x)11×1
(1+x)
= (1+x)11 x11
Coefficientofx7is11C7=11C11 7=11C4=330
[ ( ) ]
( )
7
n=1
n(n+1)(2n +1)
4
1
4
7
n=1
(2n3+3n2+n)
=1
427.8
2
2+37.8.15
6+7.8
2
1
4[2×49 ×16 +28 ×15 +28]
=1
4[1568 +420 +28] = 504
[ ] [ ]
[ ( ) ( ) ]
Question147
Thesum 20
k=1(1+2+3+ ... + k)is
[Jan.8,2020(I)]
Answer:1540
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question148
Ifthesumofthefirst40termsoftheseries,3 +4+8+9
+13 +14 +18 +19 + ...is(102)m,thenmisequalto:
[Jan.7,2020(II)]
Options:
A.20
B.25
C.5
D.10
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question149
Ifthesumoffirst11termsofanA.P.,a1,a2,a3, ...is0(a10),thenthesum
oftheA.P.,a1,a3,a5, ..., a23iska1,wherekisequalto:
[Sep.02,2020(II)]
Options:
A.121
10
Givenseriescanbewrittenas
20
k=1
k(k+1)
2=1
2
20
k=1
(k2+k)
=1
2
20(21)(41)
6+20(21)
2
=1
2
420 ×41
6+20 ×21
2=1
2[2870 +210] = 1540
[ ]
[ ]
S=3+4+8+9+13 +14 +18 +19..... . 40terms
S=7+17 +27 +37 +47 + ... . .20terms
S40 =20
2[2×7+ (19)10] = 10[14 +190]
=10[2040] = (102)(20)
m=20
B. 121
10
C. 72
5
D.72
5
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question150
IfthefirsttermofanA.P.is3andthesumofitsfirst25termsisequaltothe
sumofitsnext15terms,thenthecommondifferenceofthisA.P.is:
[Sep.03,2020(I)]
Options:
A. 1
6
B. 1
5
C. 1
4
D. 1
7
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question151
Inthesumoftheseries20 +19 3
5+19 1
5+18 4
5+ ...upton th termis488and
Letcommondifferencebed
S11 =011
2{2a1+10 d} = 0
a1+5d =0d= a1
5...(i)
Now,S=a1+a3+a5+ ... + a23
=a1+ (a1+2d ) + (a1+4d ) + .... + (a1+22d )
=12a1+2d 11 ×12
2
=12 a1+11 a1
5(From(i))
=12 × 6
5a1= 72
5a1
[ ( ) ]
( )
Givena=3andS25 =S40 S25
2S25 =S40
2×25
2[6+24d ] = 40
2[6+39d ]
25[6+24d ] = 20[6+39d ]
5(2+8d ) = 4(2+13d )
10 +40d =8+52d
d=1
6
thenn th termisnegative,then:
[Sep.03,2020(II)]
Options:
A.n =60
B.n th termis-4
C.n =41
D.n th termis42
5
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question152
Leta1,a2, ...., anbeagivenA.P.whosecommondifferenceisanintegerand
Sn=a1+a2+ .... + an.Ifa1=1,an=300and15 n50,thentheordered
pair(Sn4,an4)isequalto:
[Sep.04,2020(II)]
Options:
A.(2490,249)
B.(2480,249)
C.(2480,248)
D.(2490,248)
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Sn=20 +19 3
5+19 1
5+18 4
5+ ....
Sn=488
488 =n
22100
5+ (n1) 2
5488 =n
2(101 n) n2101n +2440 =0
n=61or40
Forn=40 Tn>0
Forn=61 Tn<0
nth term = T61 =100
5+ (61 1) 2
5= 4
[ ( ) ( ) ]
( )
Giventhata1=1andan=300anddZ
300 =1+ (n1)d
d=299
(n1)=23 ×13
(n1)
disaninteger
n1=13or23
n=14or24 (∵15 n50)
n=24andd=13
a20 =1+19 ×13 =248
s20 =20
2(2+19 ×13) = 2490
Question153
If32 sin 2 α 1,14and342 sin 2 αarethefirstthreetermsofanA.P.forsomeα,
thenthesixthtermofthisA.Pis:
[Sep.05,2020(I)]
Options:
A.66
B.81
C.65
D.78
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question154
Ifthesumofthefirst20termsoftheseries
log(712)x+log(713)x+log(714)x+ ...is460,thenxisequalto:
[Sep.05,2020(II)]
Options:
A.72
B.7
1
2
C.e2
D.7
46
21
Answer:A
Solution:
Solution:
Giventhat32 sin 2 α 1,14,342 sin 2 αareinA.P.
So,32 sin 2 α 1+342 sin 2 α =28
32 sin 2 α
3+81
32 sin 2 α =28
Let32 sin 2 α =x
x
3+81
x=28
x284x +243 =0x=81,x=3
Whenx=81 sin 2 α =2(Notpossible)
Whenx=3α=π
12
a=30=1,d=14 1=13
a6=a+5d =1+65 =66
S=log7x2+log7x3+log7x4+ .. . 20terms
S=460
log7(x2x3x4⋅..... . x21) = 460
log7x(2+3+4..... . 21)=460
(2+3+4+ ..... + 21)log7x=460
-------------------------------------------------------------------------------------------------
Question155
Iff (x+y) = f(x)f(y)and
x=1f(x) = 2,x,yN,whereN isthesetofallnatural
numbers,thenthevalueof f(4)
f(2)is:
[Sep.06,2020(I)]
Options:
A. 2
3
B. 1
9
C. 1
3
D. 4
9
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question156
Leta,b,c,d andpbeanynonzerodistinctrealnumberssuchthat
(a2+b2+c2)p22(ab +bc +cd )p+ ( b2+c2+d 2) = 0.Then:
[Sep.06,2020(I)]
Options:
A.a,c,pareinA.P.
B.a,c,pareinG.P.
C.a,b,c,d areinG.P.
D.a,b,c,d areinA.P.
Answer:C
Solution:
Solution:
20
2(2+21)log7x=460
log7x=460
230 =2x=72=49
Letf (1) = k,thenf(2) = f(1+1) = k2
f(3) = f(2+1) = k3
x=1
f(x) = 2k+k2+k3+ ... . =2
k
1k=2k=2
3
Now, f(4)
f(2)=k4
k2=k2=4
9.
Rearrangegivenequation,weget
(a2p22abp +b2) + (b2p22bcp +c2)
+(c2p22cd p +d2) = 0
-------------------------------------------------------------------------------------------------
Question157
ThecommondifferenceoftheA.P.b1,b2, ..., bmis2morethanthecommon
differenceofA.P.a1,a2, ..., an.Ifa40 = 159,a100 = 399andb100 =a70,then
b1isequalto:
[Sep.06,2020(II)]
Options:
A.81
B.-127
C.-81
D.127
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question158
Supposethatafunctionf :RRsatisfiesf (x+y) = f(x)f(y)forallx,yR
andf (a) = 3.If
i=1
nf(i) = 363,thennisequalto_______.
[NASep.06,2020(II)]
Answer:5
Solution:
(ap b)2+ (bp c)2+ (cp d)2=0
ap b=bp c=cp d=0
b
a=c
b=d
ca,b,c,dareinG.P.
Letcommondifferenceofseries
a1,a2,a3, ......, anbed
a40 =a1+39d = 159...(i)
anda100 =a1+99d = 399...(ii)
Fromequations(i)and(ii),
d= 4anda1= 3
Since,thecommondifferenceofb1,b2, ......, bnis2morethancommondifferenceofa1,a2, ......, an
∴Commondifferenceofb1,b2,b3, .....is(-2)
b100 =a70
b1+99(−2) = (−3) + 69(−4)
b1=198 279 b1= 81
f(x+y) = f(x) f(y) xRandf(1) = 3
f(x) = 3xf(i) = 3i
n
i=1
f(i) = 363
3+32+33+ .... + 3n=363
-------------------------------------------------------------------------------------------------
Question159
If210 +2931+2832+ .... + 2×39+310 =S211thenSisequalto:
[Sep.05,2020(I)]
Options:
A.311 212
B.311
C. 311
2+210
D.2 311
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question160
Ifthesumofthesecond,thirdandfourthtermsofapositivetermG.P.is3
andthesumofitssixth,seventhandeighthtermsis243,thenthesumofthe
first50termsofthisG.P.is:
[Sep.05,2020(II)]
Options:
A. 1
26(349 1)
B. 1
26(350 1)
C. 2
13(350 1)
D. 1
13(350 1)
Answer:B
Solution:
3(3n1)
31=363 Sn=a(rn1)
(r1)
3n1=363 ×2
3=242
3n=243 =35n=5
[ ]
GivensequenceareinG.P.andcommonratio3
210 3
2
11 1
3
21
=S211
210
311 211
211
1
2
=S211
311 211 =S211 S=311
( ( ) )
( )
( )
:
-------------------------------------------------------------------------------------------------
Question161
Letαandβbetherootsofx23x +p=0andγandδbetherootsof
x26x +q=0.Ifα,β,γ,δformageometricprogression.Thenratio
(2q +p) : (2q p)is:
[Sep.04,2020(I)]
Options:
A.3:1
B.9:7
C.5:3
D.33:31
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question162
Thevalueof(0.16)log2.5
1
3+1
32+1
33+ ...to∞ )isequalto________.
[NASep.03,2020(I)]
Answer:4
Solution:
(
Letthefirsttermbe'a'andcommonratiobe'r'.
ar(1+r+r2) = 3...(i)
andar5(1+r+r2) = 243...(ii)
From(i)and(ii),
r4=81 r=3anda=1
13
S50 =a(r50 1)
r1=350 1
26 Sn=a(rn1)
(r1)
[ ]
Letα,β,γ,δbeinG.P.,thenαδ =βγ
α
β=γ
δαβ
α+β= γδ
γ+δ
94p
3=36 4q
6
36 16p =36 4q q=4p
2q +p
2q p=8p +p
8p p=9p
7p =9
7
| |
(0.16)
log2.5
1
3+1
32+1
33+ .... .
0.16
log2.5
1
3
11
3S=a
1r
( )
( ) [ ]
-------------------------------------------------------------------------------------------------
Question163
ThesumofthefirstthreetermsofaG.P.isSandtheirproductis27.Then
allsuchSliein:
[Sep.02,2020(I)]
Options:
A.(−, 9] [3,)
B.[−3,)
C.(−, 3] [9,)
D.(−,9]
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question164
If|x| < 1, | y| <1andx y,thenthesumtoinfinityofthefollowingseries
(x+y) + (x2+xy +y2) + (x3+x2y+xy2+y3) + ....is:
[Sep.02,2020(I)]
Options:
A. x+yxy
(1+x)(1+y)
B. x+y+xy
(1+x)(1+y)
C. x+yxy
(1x)(1y)
D. x+y+xy
(1x)(1y)
Answer:C
=0.16
log2.5
1
2
= (2.5)2log2.5
1
2=1
2
2=4
( )
( ) ( )
LettermsofG.P.be a
r,a,ar
a1
r+1+r=S...(i)
anda3=27
a=3...(ii)
Puta=3ineqn.(1),weget
S=3+3 r +1
r
Iff(x) = x+1
x,thenf(x) (−, 2] [2,)
3f (x) (−, 6] [6,)
3+3f (x) (−, 3] [9,)
Then,itconcludesthat
S (−, 3] [9,)
( )
( )
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question165
LetSbethesumofthefirst9termsoftheseries:
{x+ka} + {x2+ (k+2)a} + {x3+ (k+4)a}+{x4+ (k+6)a} + ... where a 0
andx 1.IfS =x10 x+45a(x1)
x1,thenkisequalto:
[Sep.02,2020(II)]
Options:
A.-5
B.1
C.-3
D.3
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question166
Ifmarithmeticmeans(A.Ms)andthreegeometricmeans(G.Ms)areinserted
between3and243suchthat4 th A.M.isequalto2 nd G.M.,thenmisequal
to_______
[Sep.03,2020(II)]
Answer:39
Solution:
S= (x+y) + (x2+y2+xy) + (x3+x2y+xy2+y3) + .. .
=1
xy[ (x2y2) + (x3y3) + (x4y4+ .. . ]
=1
xy
x2
1xy2
1y=(xy)(x+yxy)
(xy)(1x)(1y)
S=a
1r
=x+yxy
(1x)(1y)
[ ]
[ ]
S= ( x+x2+x3+ .. . 9terms)
+a[k+ (k+2)+ + (k+4) + .. . 9terms]⇒S=x(x91)
x1+9
2[2ak +8× (2a)]
S=x10 x
x1+9a(k+8)
1=x10 x+45a(x1)
x1Given)
x10 x+9a(k+8)(x1)
x1=x10 x+45a(x1)
x1
9a(k+8) = 45a k+8=5k= 3
(
LetmarithmeticmeanbeA1,A2...AmandG1,G2,G3begeometricmean.
-------------------------------------------------------------------------------------------------
Question167
If1 + (1221) + (1423) + (1625) + ...... + (1202)19) =α220β,
thenanorderedpair(α,β)isequalto:
[Sep.04,2020(I)]
Options:
A.(10,97)
B.(11,103)
C.(10,103)
D.(11,97)
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question168
Letf :RRbeafunctionwhichsatisfiesf (x+y) = f(x) + f(y), x,yR.If
f(a) = 2andg(n) = (n1)
k=1f(k), nN,thenthevalueofn,forwhichg(n) = 20,
is:
[Sep.02,2020(II)]
Options:
A.5
B.20
TheA.P.formedbyarithmeticmeanis,
3,A1,A2,A3, ..... . Am,243
d=243 3
m+1=240
m+1
TheG.P.formedbygeometricmean
3,G1,G2,G3,243
r=243
3
1
3+1= (81)14=3
A4=G2
3+4240
m+1=3(3)2
3+960
m+1=27 m+1=40 m=39.
( )
( )
Thegivenseriesis
1+ (1221) + (1423) + (1625) + .. . (120219)
S=1+
10
r=1
[1 (2r)2(2r 1)]
=1+
10
r=1
(18r3+4r2) = 1+10
10
r=1
(8r34r2)
=11 810 ×11
2
2+4×10 ×11 ×21
6
=11 2× (110)2+4×55 ×7
=11 220(110 7)
=11 220 ×103 =α220β
α=11,β=103
(α,β) = (11,103)
( ) ( )
C.4
D.9
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question169
Leta,bandcbethe7 th ,11 th and13 th termsrespectivelyofanon-
constantA.P.IfthesearealsothethreeconsecutivetermsofaG.P.,then a
cis
equalto:
[Jan.09,2019(II)]
Options:
A.2
B. 1
2
C. 7
13
D.4
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question170
Ifa,bandcbethreedistinctrealnumbersinG.P.anda +b+c=xb,thenx
Given:f(x+y) = f(x) + f(y), x,yR,f(1) = 2
f(2) = f(1) + f(1) = 2+2=4
f(3) = f(1) + f(2) = 2+4=6
f(n1) = 2(n1)
Now,g(n) =
n1
k=1
f(k)
=f(1) + f(2) + f(3) + ... . f(n1)
=2+4+6+ ..... + 2(n1)
=2[1+2+3+ ...... + (n1)]
=2×(n1)(n)
2=n2n
g(n) = 20(given)
So,n2n=20
n2n20 =0
(n5)(n+4) = 0
n=5orn= 4(notpossible)
LetfirsttermandcommondifferencebeAandDrespectively.
a=A+6D,b=A+10Dandc=A+12D
Since,a,b,careinG.P.
Hence,relationbetweena,bandcis,
b2=a.c.
(A+10D)2= (A+6D)(A+12D)
14D +A=0
A= 14D
a= 8D,b= 4Dandc= 2D
a
c=8D
2D =4
cannotbe:
[Jan.09,2019(I)]
Options:
A.-2
B.-3
C.4
D.2
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question171
Thesumofthefollowingseries1 +6+9(12+22+32)
7+12(12+22+32+42)
9
+15(12+22+ ... + 52)
11 + ...upto15terms,is:
[Jan.09,2019(II)]
Options:
A.7520
B.7510
C.7830
D.7820
Answer:D
Solution:
Solution:
a,b,c,areinG.P.
b2=ac
Since,a+b+c=xb
a+c= (x1)b
Takesquareonbothsides,weget
a2+c2+2ac = (x1)2b2
a2+c2= (x1)2ac 2ac[∵b2=ac]
a2+c2=ac[(x1)22]
a2+c2=ac[x22x 1]
a2+c2arepositiveandb2=acwhichisalsopositive.
Then,x22x 1wouldbepositivebutforx=2,x22x 1isnegative.
Hence,xcannotbetakenas2.
S=1+6+9(12+22+32)
7+12(12+22+32+42)
9+ 15(12+22+32+42+52)
11 + ...
S=3 (1)2
3+6 (12+22)
5+9 (12+22+32)
7+ 12 (12+22+32+42)
9+ ...
Now,nth termoftheseries,
tn=3n (12+22+ ... + n2)
(2n +1)
tn=3n n(n+1)(2n +1)
6(2n +1)=n3+n2
2
Sn=Σtn=1
2
n(n+1)
2
2+n(n+1)(2n +1)
6
=n(n+1)
4
n(n+1)
2+2n +1
3
{ ( ) }
( )
-------------------------------------------------------------------------------------------------
Question172
Thesumofallvaluesofθ 0,π
2satisfyingsin2 +cos4 =3
4is:
[Jan.10,2019(I)]
Options:
A.π
B.
4
C. π
2
D.
8
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question173
Letαandβbetherootsofthequadraticequation
x2sin θ x(sin θ cos θ +1) + cos θ =0(0<θ<45),andα <β.Then
n=0
αn+(−1)n
βnisequalto:
[Jan.11,2019(II)]
Options:
A. 1
1cos θ 1
1+sin θ
( )
( )
Hence,sumoftheseriesupto15termsis,
S15 =15 ×16
4
15 16
2+31
3
=60 ×120 +60×31
3
=7200 +620 =7820
{ }
sin2 +cos4 =3
4
1cos2 +cos4 =3
4
cos2(1cos2) = 1
4...(i)
∵G.M.≤A.M.
(cos2)(1cos2) cos2 + ( 1cos2
2
2
=1
4...(ii)
So,fromequation(i)and(ii),weget.
G.M.=A.M.
Itispossibleonlyif
cos2 =1cos2
cos2 =1
2cos 2 θ = ± 1
2
θ=π
8,
8∴Sum = π
8+
8=π
2
( )
B. 1
1+cos θ +1
1sin θ
C. 1
1cos θ +1
1+sin θ
D. 1
1+cos θ 1
1sin θ
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question174
Leta1,a2, ..., a10beaG.P.If a3
a1=25,then a9
a5equals:
[Jan.11,2019(I)]
Options:
A.54
B.4(52)
C.53
D.2(52)
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question175
x2sin θ x(sin θ cos θ +1) + cos θ =0.
x2sin θ x sin θ cos θ x+cos θ =0
x sin θ(xcos θ) 1(xcos θ) = 0
(xcos θ)(x sin θ 1) = 0
x=cos θ,cosecθ,θ (0,45)
α=cos θ,β=cosecθ
n=0
αn=1+cos θ +cos2θ+ .. . =1
1cos θ
n=0
(−1)n
βn=11
cosecθ +1
cosec2θ1
cosec3θ+ .. .
=1sin θ +sin2θsin3θ+ .. .
=1
1+sin θ
n=0
αn+(−1)n
βn=
n=0
αn+
n=0
(−1)n
βn
=1
1cos θ +1
1+sin θ
( )
Leta1=a,a2=ar,a3=ar2...a10 =ar9
wherer= commonratioofgivenG.P.
Given, a3
a1
=25
ar2
a=25
r= ±5
Now, a9
a5
=ar8
ar4=r4= 5)4=54
Thesumofaninfinitegeometricserieswithpositivetermsis3andthesum
ofthecubesofitstermsis 27
19.Thenthecommonratioofthisseriesis:
[Jan.11,2019(I)]
Options:
A. 1
3
B. 2
3
C. 2
9
D. 4
9
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question176
LetSn=1+q+q2+ .... + qnandT n=1+q+1
2+q+1
2
2+ ... + q+1
2
n
whereqisarealnumberandq 1.If
101C1+101C2S1+ .... + 101C101 S100 =αT 100,thenαisequalto:
[Jan.11,2019(II)]
Options:
A.299
B.202
C.200
D.2100
Answer:D
Solution:
( ) ( ) ( )
Letthetermsofinfiniteseriesarea,ar,ar2,ar3, ...
So, a
1r=3
Since,sumofcubesofitstermsis 27
19thatissumofa3,
a3r3, .. . is 27
19
So, a3
1r3=27
19
a
1r×a2
(1+r2+r)=27
19
9(1+r22r) × 3
1+r2+r=27
19
6r213r +6=0
(3r 2)(2r 3) = 0
r=2
3,or 3
2
As|r| < 1
So,r=2
3
-------------------------------------------------------------------------------------------------
Question177
TheproductofthreeconsecutivetermsofaG.P.is512.If4isaddedtoeach
ofthefirstandthesecondoftheseterms,thethreetermsnowformanA.P.
ThenthesumoftheoriginalthreetermsofthegivenG.P.is:
[Jan.12,2019(I)]
Options:
A.36
B.32
C.24
D.28
Answer:D
Solution:
-------------------------------------------------------------------------------------------------
Sn=1qn+1
1q,Tn=
1q+1
2
n+1
1q+1
2
T100 =
1q+1
2
101
1q+1
2
Sn =1
1qqn+1
1q,T100 =2101 (q+1)101
2100(1q)
Now,101C1+101C2S1+101C3S2+ ... + 101C101S100
=1
1q(101C2+ ... + 101C101)
1
1q(101C2q2+101C3q3+ ... + 101C101q101) + 101
=1
1q(2101 1101) 1
1q( (1+q)101 1
101C1q) + 101
=1
1q[2101 102 (1+q)101 +1+101q] + 101
=1
1q[2101 101 +101q (1+q)101] + 101
=1
1q[2101 (1+q)101] = 2100T100
Hence,bycomparisonα=2100
( ) ( )
( )
( )
( )
( )
( )
( )
LetthreetermsofaG.P.be a
r,a,ar
a
raar =512
a3=512 a=8
4isaddedtoeachofthefirstandthesecondofthreetermsthenthreetermsare, 8
r+4,8+4,8r.
8
r+4,12,8rformanA.P.
2×12 =8
r+8r +4
2r25r +2=0
(2r 1)(r2) = 0
r=1
2or2
Therefore,sumofthreeterms = 8
2+8+16 =28
Question178
LetSk=1+2+3+ ..... + k
kIfS1
2+S2
2+ ..... + S10
2=5
12A.ThenAisequalto
[Jan.12,2019(I)]
Options:
A.283
B.301
C.303
D.156
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question179
Ifthesumofthefirst15termsoftheseries
3
4
3+11
2
3+21
4
3+33+33
4
3+ ......isequalto225kthenkisequal
to:
[Jan.12,2019(II)]
Options:
A.108
B.27
C.54
D.9
Answer:B
Solution:
Solution:
( ) ( ) ( ) ( )
1+2+3+ ... + k=k(k+1)
2
Sk=k(k+1)
2k =k+1
2
5
12A=1
4[22+33+ ... + 112]
=1
4[12+22+ ... + 1121]
=1
4
11(11 +1)(2×11 +1)
61
1
4
11 ×12 ×23
61
=1
4[505]
A=505
4×12
5=303
[ ]
[ ]
S=3
4
3+3
2
3+9
4
3+ (3)3+ ...
S=3
4
3+6
4
3+9
4
3+12
4
3+ ...
LetthegeneraltermofSbe
( ) ( ) ( )
( ) ( ) ( ) ( )
-------------------------------------------------------------------------------------------------
Question180
I f nC4,nC5andnC6areinA.P.,thenncanbe
[Jan.12,2019(II)]
Options:
A.9
B.14
C.11
D.12
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question181
If19thtermofanon-zeroA.P.iszero,thenits(49thterm):(29thterm)is:
[Jan.11,2019(II)]
Options:
A.4:1
B.1:3
C.3:1
D.2:1
Answer:C
Solution:
Solution:
Tr=3r
4
3,then
255K =
15
r=1
Tr=3
4
315
r=1
r3
255K =27
64 ×15 ×16
2
2
K=27
( )
( )
( )
SincenC4,nC5andnC6areinA.P.
2nC5=nC4+nC6
2=
nC4
nC5
+
nC6
nC5
2=5
n4+n5
5
12(n4) = 30 +n29n +20
n221n +98 =0
(n7)(n14) = 0
(n7)(n14) = 0
n=7,n=14
LetfirsttermandcommondifferenceofAPbeaanddrespectively,then
tn=a+ (n1)d
t19 =a+18d =0∴ a= 18d
-------------------------------------------------------------------------------------------------
Question182
Thesumofalltwodigitpositivenumberswhichwhendividedby7yield2or
5asremainderis:
[Jan.10,2019(I)]
Options:
A.1256
B.1465
C.1365
D.1356
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question183
Leta1,a2, ......, a30beanA.P.,S =30
i=1aiandT =15
i=1a(2i 1)Ifa5=27and
S2T =75,thena10isequalto:
[Jan.09,2019(I)]
Options:
A.52
B.57
C.47
D.42
Answer:A
Solution:
Solution:
t49
t29
=a+48d
a+28d
=18d +48d
18d +28d =30d
10d =3
t49 :t29 =3:1
Twodigitpositivenumberswhichwhendividedby7yield2asremainderare12termsi.e,16,23,30, ..., 93
Twodigitpositivenumberswhichwhendividedby7yield
5asremainderare13termsi.e,12,19,26, ..., 96
ByusingAPsumof16,23, ..., 93,weget
S1=16 +23 +30 + ... + 93 =654
ByusingAPsumof12,19,26, ..., 96,weget
S1=12 +19 +26 + ... + 96 =702
∴requiredSum =S1+S2=654 +702 =1356
S=
30
i=1
ai=30
2[2a1+29d ]
T=
15
i=1
a(2i 1)=15
2[2a1+28d ]
Since,S2T =75
-------------------------------------------------------------------------------------------------
Question184
Ifthreedistinctnumbersa,b,careinG.P.andtheequations
ax2+2bx +c=0andd x2+2ex +f=0haveacommonroot,thenwhichone
ofthefollowingstatementsiscorrect?
[April08,2019(II)]
Options:
A. d
a,e
b,f
careinA.P.
B.d ,e,f areinA.P.
C.d ,e,f areinG.P.
D. d
a,e
b,f
careinG.P.
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question185
Forx R,let[x]denotethegreatestintegerx,thenthesumoftheseries
1
3+ 1
31
100 + 1
32
100 + s+ 1
399
100 is
[April12,2019(I)]
Options:
A.-153
B.-133
C.-131
D.-135
Answer:B
[ ] [ ] [ ] [ ]
30a1+435d 30a1420d =75
d=5
Also,a5=27 a1+4d =27 a1=7
Hence,a10 =a1+9d =7+9×5=52
Sincea,b,careinG.P.
b2=acc
Givenequationis,ax2+2bx +c=0
ax2+2acx +c=0 (√ax + c)2=0
x= c
a
Also,giventhatax2+2bx +c=0andd x2+2ex +f=0haveacommonroot.
x= c
amustsatisfyd x2+2ex +f=0
dc
a+2e c
a+f=0
d
a2e
ac +f
c=0d
a2e
b+f
c=0
2e
b=d
a+f
cd
a,e
b,f
careinA.P.
()
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question186
Thesum 3×13
12+5× (13+23)
12+22+7× (13+23+33)
12+22+32+ ......upto10 th term,is:
[April10,2019(I)]
Options:
A.680
B.600
C.660
D.620
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question187
Thesum1 +13+23
1+2+13+23+33
1+2+3+ ......+13+23+33+ ... + 153
1+2+3+ ... + 15 1
2(1+2+3+ ... + 15is
equalto:
[April10,2019(II)]
Options:
A.620
B.1240
C.1860
D.660
)
[x] + x+1
n+x+2
n... . x+n1
n= [nx]
and[x] + [−x] = 1(xz)
1
3+ 1
3100 + ... + 1
399
100
= 100 1
3+1
3+1
100 + .. . 1
3+99
100
= 100 100
3= 133
[ ] [ ] [ ]
[ ] [ ] [ ]
{ [ ] [ ] [ ] }
[ ]
rth termoftheseries,
Tr=(2r +1)(13+23+33+ ... + r3)
12+22+32+ ... + r2
Tr= (2r +1)r(r+1)
2
2×6
r(r+1)(2r +1)=3r(r+1)
2
∴sumof10termsis = S=
10
r=1
Tr=3
2
10
r=1
(r2+r)
=3
2
10 × (10 +1)(2×10 +1)
6+10 ×11
2
=3
2×5×11 ×8=660
( )
{ }
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question188
Thesumoftheseries1 +2×3+3×5+4×7+ ......upto11 th termis:
[April09,2019(II)]
Options:
A.915
B.946
C.945
D.916
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question189
Someidenticalballsarearrangedinrowstoformanequilateraltriangle.The
firstrowconsistsofoneball,thesecondrowconsistsoftwoballsandsoon.
If99moreidenticalballsareaddedtothetotalnumberofballsusedin
formingtheequilateraltriangle,thenalltheseballscanbearrangedina
squarewhoseeachsidecontainsexactly2ballslessthanthenumberofballs
eachsideofthetrianglecontains.Thenthenumberofballsusedtoformthe
equilateraltriangleis:
[April09,2019(II)]
Options:
A.157
Let,S=1+13+23
1+2+13+23+33
1+2+3+ .. . 15terms
Tn=13+23+ .. . n3
1+2+ .. . n=
n(n+1)
2
2
n(n+1)
2
=n(n+1)
2
Now,S=1
2
15
n=1
n2+
15
n=1
n=1
2
15(16)(31)
6+15(16)
2
=680
∴requiredsumis,680 1
2
15(16)
2=680 60 =620
( )
( ) ( )
1+2.3 +3.5 +4.7 + .......Let,S= (2.3 +3.5 +4.7 + ......)
Now,S10 =
10
n=1
(n+1)(2n +1) =
10
n=1
(2n2+3n +1)
=2n(n+1)(2n +1)
6+3n(n+1)
2+n
Putn=10
=2.10.11.21
6+3.10.11
2+10 =945
Hencerequiredsumoftheseries = 1+945 =946
B.262
C.225
D.190
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question190
Thesum 20
k=1k1
2kisequalto:
[April08,2019(II)]
Options:
A.2 3
217
B.1 11
220
C.2 11
219
D.2 21
220
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Numberofballsusedinequilateraltriangle
=n(n+1)
2
∵sideofequilateraltrianglehasn-balls
∴no.ofballsineachsideofsquareis = (n2)
Accordingtothequestion, n(n+1)
2+99 = (n2)2
n2+n+198 =2n28n +8
n29n 190 =0 (n19)(n+10) = 0
n=19
Numberofballsusedtoformtriangle
=n(n+1)
2=19 ×20
2=190
Let,S=
20
k=1
k1
2k
S=1
2+21
22+31
23+ .... + 201
220...(i)
1
2S=1
22+21
23+ .... + 19 1
220 +20 1
221...(ii)
Onsubtractingequations(ii)by(i),
S
2=1
2+1
22+1
23+ .... + 1
220 20 1
221
=
1
211
220
11
2
201
221 =11
220 101
220
S
2=1111
220 S=2111
219 =211
219
( )
( )
Question191
LetSndenotethesumofthefirstntermsofanA.P.IfS4=16andS6= 48,
thenS10isequalto:
[April12,2019(I)]
Options:
A.-260
B.-410
C.-320
D.-380
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question192
Ifa1,a2,a3, ......areinA.P.suchthata1+a7+a16 =40,thenthesumofthe
first15termsofthisA.P.is:
[April12,2019(II)]
Options:
A.200
B.280
C.120
D.150
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Given,S4=16andS6= 48
2(2a +3d ) = 16 2a +3d =8...(i)
And3[2a +5d ] = 48 2a +5d = 16
2d = 24 [usingequation(i)]
d= 12anda=22
S10
10
2= (44 +9(−12)) = 320
LetthecommondifferenceoftheA.P.is'd'.
Given,a1+a7+a16 =40
a1+a1+6d +a1+15d =40
3a1+21d =40
a1+7d =40
3
Now,sumoffirst15termsofthisA.P.is,
S15 =15
2[2a1+14d ] = 15(a1+7d )
=15 40
3=200
[Using(i)]
( )
Question193
Ifa1,a2,a3, .... . anareinA.P.anda1+a4+a7+ .... + a16 =114then
a1+a6+a11 +a16isequalto:
[April10,2019(I)]
Options:
A.98
B.76
C.38
D.64
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question194
Letthesumofthefirstntermsofanon-constantA.P.
a1,a2,a3, ..............be50n +n(n7)
2A,whereAisaconstant.Ifd isthe
commondifferenceofthisA.P.,thentheorderedpair(d,a50)isequalto:
[April09,2019(I)]
Options:
A.(50,50 +46A)
B.(50,50 +45A)
C.(A,50 +45A)
D.(A,50 +46A)
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question195
Let
k=1
10 f(a+k) = 16(210 1),wherethefunctionf satisfies
a1+a4+a7+ ... + a16 =114
3(a1+a16) = 114
a1+a16 =38
Now,a1+a6+a11 +a16 =2(a1+a16) = 2×38 =76
Sn=50 7A
2n+n2×A
2a1=50 3S
d=a2a1=Sn2
Sn1
Sn1
d=A
2×2=A
Now,a50 =a1+49 ×d
= (50 3A) + 49A =50 +46A
So,(d,a50) = (A,50 +46A)
( )
f(x+y) = f(x)f(y)forallnaturalnumbersx,yandf (a) = 2Thenthenatural
number'a'is:
[April09,2019(I)]
Options:
A.2
B.16
C.4
D.3
Answer:D
Solution:
Question196
IfthesumandproductofthefirstthreetermsinanA.P.are33and1155,
respectively,thenavalueofits11 th termis:
[April09,2019(II)]
Options:
A.-35
B.25
C.-36
D.-25
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
f(x+y) = f(x) f(y)
Letf (x) = tx
f(1) = 2
t=2
f(x) = 2x
Since,
10
k=1
f(a+k) = 16(210 1)
Then,
10
k=1
2a+k=16(210 1)
2a10
k=1
2k=16(210 1)
2a×((210) 1) × 2
(21)=16 × (210 1)2.2a=16
a=3
LetthreetermsofA.P.aread,a,a+d
Sumoftermsis,ad+a+a+d=33 a=11
Productoftermsis,(ad)a(a+d) = 11(121 d2) = 1155
121 d2=105 d= ±4ifd=4
T11 =T1+10d =7+10(4) = 47
ifd= 4
T11 =T1+10d =15 +10(−4) = 25
Question197
Thesumofallnaturalnumbers'n'suchthat100 <n<200andH.C.F.
(91,n) > 1is:
[April08,2019(I)]
Options:
A.3203
B.3303
C.3221
D.3121
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question198
Ifα,βandγarethreeconsecutivetermsofanonconstantG.P.suchthatthe
equationsαx2+2βx +γ=0andx2+x1=0haveacommonroot,then
α(β+γ)isequalto:
[April12,2019(II)]
Options:
A.0
B.αβ
C.αγ
D.βγ
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
91 =13 ×7
Then,therequirednumbersareeitherdivisibleby7or13.
∴Sumofsuchnumbers=Sumofno.divisibleby7+
sumoftheno.divisibleby13−Sumofthenumbersdivisibleby91
= (105 +112 + ... + 196) + (104 +117 + ... + 195) 182
=2107 +1196 182 =3121
α,β,γarethreeconsecutivetermsofanon-constantG.P.
β2=αγ
Sorootsoftheequationαx2+2βx +γ=0are
±2β2αγ
=β
α
αx2+2βx +γ=0andx2+x1=0haveacommonroot.
∴thisrootsatisfytheequationx2+x1=0
β2αβ α2=0
αγ αβ α2=0α+β=γ
Now,α(β+γ) = αβ +αγ
=αβ +β2= (α+β)β=βγ
Question199
Leta,bandcbeinG.P.withcommonratior,wherea 0and0 <r1
2.If
3a,7band15carethefirstthreetermsofanA.P.,thenthe4 th termofthis
A.P.is:
[April10,2019(II)]
Options:
A. 2
3a
B.5a
C. 7
3a
D.a
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question200
Let 1
x1
,1
x2
,1
x3
, ...., ( xi0fori =1,2, ..., n)beinA.P.
suchthatx1=4andx21 =20.Ifnistheleastpositiveintegerforwhich
xn>50,then n
i=1
1
xi
isequalto.
[OnlineApril16,2018]
Options:
A.3
B. 13
8
C. 13
4
D. 1
8
Answer:C
Solution:
( )
a,b,careinG.P.⇒b=ar,c=ar2
3a,7b,15careinA.P.
3a,7ar,15ar2areinA.P.
14ar =3a +15ar2
15r214r +3=0r=1
3or 3
5
r<1
2r=3
5rejected
Fourthterm = 15ar2+7ar 3a
=a(15r2+7r 3) = a15
9+7
33=a
( )
-------------------------------------------------------------------------------------------------
Question201
Ifx1,x2, ...., xnand 1
h1
,1
h2, ...⋅ 1
hn
aretwoA.P'ssuchthatx3=h2=8and
x8=h7=20,thenx5.h10equals.
[OnlineApril15,2018]
Options:
A.2560
B.2650
C.3200
D.1600
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question202
1
x1
,1
x2
,1
x3
, ...., 1
xn
areinA.P
x1=4andx21 =20
Letd'bethecommondifferenceofthisA.P
its21 st term = 1
x21
=1
x1
+ [(21 1) × d]
d=1
20 ×1
20 1
4d= 1
100
Also xn>50(given).
1
xn
=1
x1
+ [(n1) × d]
xn=x1
1+ (n1) × d×x1
x1
1+ (n1) × d×x1
>50
4
1+ (n1) × 1
100 ×4
>50
1+ (n1) × 1
100 ×4<4
50
1
100(n1) < 23
100
n1>23 n>24
Therefore,n=25.
25
i=1
1
xi
=25
22×1
4+ (25 1) × 1
100 =13
4
( )
( )
( )
[ ( ) ( ) ]
Supposed1isthecommondifferenceoftheA.P.x1,x2, ... . xnthen
x8x3=5d 1=12 d1=12
5=2.4
x5=x3+2d 1=8+2×12
5=12.8
Supposed2isthecommondifferenceoftheA.P
1
h1
,1
h2
, ...⋅ 1
hn
then
5d 2=1
20 1
8=3
40 d2=3
200
1
h10
=1
h7
+3d 2=1
200 h10 =200
x5h10 =12.8 ×200 =2560
IfbisthefirsttermofaninfiniteG.Pwhosesumisfive,thenbliesinthe
interval.
[OnlineApril15,2018]
Options:
A.(−, 10)
B.(10,)
C.(0,10)
D.(-10,0)
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question203
LetAn=3
43
4
2+3
4
3 ... + (−1)n13
4
nandBn=1An.Then,the
leastoddnaturalnumberp,sothatBn>An,foralln pis
[OnlineApril15,2018]
Options:
A.5
B.7
C.11
D.9
Answer:B
Solution:
Solution:
( ) ( ) ( ) ( )
Firstterm = bandcommonratio = r
Forinfiniteseries,Sum =b
1r=5
b=5(1r)
So,intervalofb= (0,10)as,−1<r<1forinfiniteG.P.
An=3
43
4
2+3
4
3 ... + (−1)n13
4
n
WhichisaG.P.witha=3
4,r=3
4andnumberofterms = n
An=
3
4×13
4
n
13
4
=
3
4×13
4
n
7
4
An=3
713
4
n...(i)
As,Bn=1An
Forleastoddnaturalnumberp,suchthatBn>An
1An>An1>2×AnAn<1
2
Fromeqn.(i),weget
3
7×13
4
n<1
213
4
n<7
6
17
6<3
4
n1
6<3
4
n
( ) ( ) ( ) ( )
( ( ) )
( )
( ( ) )
[ ( ) ]
[ ( ) ] ( )
( ) ( )
-------------------------------------------------------------------------------------------------
Question204
Ifa,b,careinA.P.anda2,b2,c2areinG.P.suchthata <b<cand
a+b+c=3
4,thenthevalueofais
[OnlineApril15,2018]
Options:
A. 1
41
32
B. 1
41
42
C. 1
41
2
D. 1
41
22
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question205
Thesumofthefirst20termsoftheseries1 +3
2+7
4+15
8+31
16 + ...is?
[OnlineApril16,2018]
Options:
A.38 +1
220
Asnisodd,then 3
4
n= 3n
4
So 1
6< 3
4
n1
6>3
4
n
log 1
6=n log 3
46.228 <n
Hence,nshouldbe7.
( )
( ) ( )
( ) ( )
a,b,careinA.P.thena+c=2b
alsoitisgiventhat,
a+b+c=3
4...(i)
2b +b=3
4b=1
4...(ii)
Againitisgiventhat,a2,b2,c2areinG.P.then
(b2)2=a2c2ac = ± 1
16...(iii)
From(i),(ii)and(iii),weget;
a±1
16a =1
216a28a ±1=0
CaseI:16a28a +1=0
a=1
4notpossibleasa<b)
CaseII:16a28a 1=0a=8± 128
32
a=1
4±1
22
a=1
41
22(∵a<b)
(
B.39 +1
219
C.39 +1
220
D.38 +1
219
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question206
LetAbethesumofthefirst20termsandBbethesumofthefirst40terms
oftheseries12+222+32+242+52+262+ ...IfB 2A =100λ,thenλis
equalto
[2018]
Options:
A.248
B.464
C.496
D.232
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question207
Thegeneraltermofthegivenseries = 2×2r1
2r,wherer0
∴req.sum =1+
19
r=1
2×2r1
2r
Now,
19
r=1
2×2r1
2r=
19
r=1
21
2r
=2(19)
1
211
2
19
11
2
=38 +
1
2
19 1
1
=38 +1
2
19 1=37 +1
2
19
∴req.sum =1+37 +1
2
19 =38 +1
2
19
( ) ( )
( ( ) ) ( )
( ) ( )
( ) ( )
Here,B2A
=
40
n=1
an2
20
n=1
an=
40
n=21
an2
20
n=1
an
B2A = (212+2.222+232+2.242+ ..... + 402)
(12+2.22+32+2.42..... + 202)
=20[22 +2.24 +26 +2.28 + ..... + 60]
=20[(22 +24 +26...60)20 terms + (24 +28 + .... + 60)10 terms ]
20 20
2(22 +60) + 10
2(24 +60)
=10[20.82 +10.84]
=100[164 +84] = 100.248
[ ]
Leta1,a2,a3, ..., a49beinA.P.suchthat 12
k=0a4k +1=416anda9+a43 =66.If
a1
2+a2
2+ ... + a17
2=140m,thenmisequalto
[2018]
Options:
A.68
B.34
C.33
D.66
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question208
Foranythreepositiverealnumbersa,bandc,
9(25a2+b2) + 25(c23ac) = 15b(3a +c).Then
[2017]
Options:
A.a,bandcareinG.P.
B.b,candaareinG.P.
C.b,candaareinA.P.
D.a,bandcareinA.P.
Answer:C
Solution:
Solution:
k=0
12 a4k +1=416 13
2[2a1+48d ] = 416
a1+24d =32
N ow,a9+a43 =66 2a1+50d =66
Fromeq.(i)&(ii)weget;d=1anda1=8
Also,
17
r=1
ar
2=
17
r=1
[8+ (r1)1]2=140m
17
r=1
(r+7)2=140m
17
r=1
(r2+14r +49) = 140m
17 ×18 ×35
6+14 17 ×18
2+ (49 ×17) = 140
m=34
( ) ( )
\Wehave
9(25a2+b2) + 25(c23ac) = 15b(3a +c)
225a2+9b2+25c275ac =45ab +15bc
(15a)2+ (3b)2+ (5c)275ac 45ab 15bc =0
1
2[(15a 3b)2+ (3b 5c)2+ (5c 15a)2] = 0
Itispossiblewhen15a 3b =0,3b 5c =0and5c 15a =0
15a =3b b=5a
-------------------------------------------------------------------------------------------------
Question209
Ifthreepositivenumbersa,bandcareinA.P.suchthatabc =8,thenthe
minimumpossiblevalueofbis:
[OnlineApril9,2017]
Options:
A.2
B.4
1
3
C.4
2
3
D.4
Answer:A
Solution:
Question210
Ifthearithmeticmeanoftwonumbersaandb,a>b>0,isfivetimestheir
geometricmean,then a+b
abisequalto:
[OnlineApril8,2017]
Options:
A. 6
2
B. 32
4
C. 73
12
D. 56
12
Answer:D
Solution:
Solution:
b=5c
3,a=c
3
a+b=c
3+5c
3=6c
3
a+b=2c
b,c,aareinA.P.
A.T.Q.
A.M.=5G.M.
ByArithmeticMean:
a+c=2b
Considera=b=c=2
abc =8
a+b=2b
∴minimumpossiblevalueofb=2
-------------------------------------------------------------------------------------------------
Question211
Leta,b,cR.Iff(x) = ax2+bx +cissuchthata +b+c=3and
f(x+y) = f(x) + f(y) + xy, x,yR,then
n=1
10 f(n)isequalto:
[2017]
Options:
A.255
B.330
C.165
D.190
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question212
LetSn=1
13+1+2
13+23+1+2+3
13+23+33+ ....+1+2+ ...... + n
13+23+ ...... + n3,If100Sn=n,thennisequal
to:
[OnlineApril9,2017]
Options:
A.199
a+b
2=5ab
a+b
ab =10
a
b=10 + 96
10 96 =10 +46
10 46
UseComponendoandDividendo
a+b
ab=20
86=5
26=56
12
f(x) = ax2+bx +c
f(1) = a+b+c=3f(1) = 3
Now f (x+y) = f(x) + f(y) + xy ...
Putx=y=1ineqn(i)
f(2) = f(1) + f(1) + 1=2f (1) + 1
f(2) = 7
f(3) = 12
Sn=3+7+12 + ....... + tn
Sn=3+7+12 + ....... + tn1+tn
0=3+4+5.........tonterm−tn
tn=3+4+5+ ...uptonterms
tn=(n2+5n)
2
Sn= tn= (n2+5n)
2
Sn=1
2
n(n+1)(2n +1)
6+5n(n+1)
2
=n(n+1)(n+8)
6
S10 =10 ×11 ×18
6=330
[ ]
B.99
C.200
D.19
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question213
Ifthesumofthefirstntermsoftheseries3+ 75 + 243 + 507 + .....is
4353,thennequals:
[OnlineApril8,2017]
Options:
A.18
B.15
C.13
D.29
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question214
Leta1,a2,a3, ..., an,beinA.P.Ifa3+a7+a11 +a15 =72,thenthesumofits
Tn=
n(n+1)
2
n(n+1)
2
2
Tn=2
n(n+1)
Sn= Tn=2
n
n=1
1
n1
n+1=2 1 1
n+1
Sn=2n
n+1
100Sn=n
100×2n
n+1=n
n+1=200
n=199
( )
( ) { }
∵√3[1+ 25 + 81 + 69 + .......] = 4353
3[1+5+9+13 + ..... + Tn] = 4353
3×n
2[2+ (n1)4] = 4353
2n +4n24n =870
4n22n 870 =0
2n2n435 =0
n=1± 1+4×2×435
4=1±59
4
n=1+59
4=15;orn=159
4=14.5
first17termsisequalto:
[OnlineApril10,2016]
Options:
A.306
B.204
C.153
D.612
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question215
Ifthe2 nd ,5th and9 th termsofanon-constantA.P.areinG.P.,thenthe
commonratioofthisG.P.is:
[2016]
Options:
A.1
B. 7
4
C. 8
5
D. 4
3
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question216
Letz =1+aibeacomplexnumber,a >0,suchthatz3isarealnumber.Then
thesum1 +z+z2+ .... + z11isequalto:
(OnlineApril10,2016)
a3+a7+a11 +a15 =72
(a3+a15) + (a7+a11) = 72
a3+a15 +a7+a11 =2(a1+a17)
a1+a17 =36
S17 =17
2[a1+a17] = 17 ×18 =306
LettheGPbea,arandar2thena=A+d;ar =A+4d ;
ar2=A+8d
ar2ar
ar a=(A+8d ) (A+4d )
(A+4d ) (A+d)
r=4
3
Options:
A.13653i
B.13653i
C.12503i
D.12503i
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question217
IfA >0,B>0andA +B=π
6,thentheminimumvalueoftan A +tan Bis:
[OnlineApril10,2016]
Options:
A.3 2
B.4 23
C. 2
3
D.2 3
Answer:B
Solution:
Solution:
z=1+ai
z2=1a2+2ai
z2z= {(1a2) + 2ai} {1+ai}
= (1a2) + 2ai + (1a2)ai 2a2
z3isreal⇒2a + (1a2)a=0
a(3a2) = 0a= 3(a>0)
1+z+z2......... . z11 =z12 1
z1=(1+ 3i)12 1
1+ 3i1
=(1+ 3i)12 1
3i
(1+ 3i)12 =212 1
2+3
2i
12
=212 cos π
3+i sin π
3
12 =212(cos 4 π +i sin 4 π) = 212
212 1
3i=4095
3i= 4095
33i= 13653i
( )
( )
tan(A+B) = tan A +tan B
1tan A tan B
1
3=y
1tan A tan Bwherey=tan A +tan B
tan A tan B =1 3y
AlsoAM GM
tan A +tan B
2 tan A tan B
y2 1 3y
y2443y
y2+43y 40
y 234ory 23+4
-------------------------------------------------------------------------------------------------
Question218
Letx,y,zbepositiverealnumberssuchthatx +y+z=12and
x3y4z5= (0.1)(600)3.Thenx3+y3+z3isequalto:
[OnlineApril9,2016]
Options:
A.342
B.216
C.258
D.270
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question219
Ifthesumofthefirsttentermsoftheseries
13
5
2+22
5
2+31
5
2+42+44
5
2+ ......,is 16
5mthenmisequalto:
[2016]
Options:
A.100
B.99
C.102
D.101
Answer:D
Solution:
( ) ( ) ( ) ( )
(y 234isnotpossibleastan A tan B >0)
x+y+z=12
AM GM
3x
3+4y
4+5z
5
12 12 x
3
3y
4
4z
5
5
x3y4z5
3344551
x3y4z5334455
x3y4z5 (0.1)(600)3
But,givenx3y4z5= (0.1)(600)3
∴allthenumberareequal
x
3=y
4=z
5( = k)
x=3k;y=4k;z=5k
x+y+z=12
3k +4k +5k =12
k=1∴ x=3;y=4;z=5
x3+y3+z3=216
( ) ( ) ( ) ( ) ( ) ( )
-------------------------------------------------------------------------------------------------
Question220
Forx R,x 1,if(1+x)2016 +x(1+x)2015 +x2
(1+x)2014 + .... + x2016 =
i=0
2016 aixi,thena17isequalto:
[OnlineApril9,2016]
Options:
A. 2017!
17!2000!
B. 2016!
17!1999!
C. 2016!
16!
D. 2017!
2000!
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question221
IfmistheA.M.oftwodistinctrealnumbersl andn(l,n>1)andG1,G2and
G3arethreegeometricmeansbetweenl andn,thenG1
4+2G2
4+G3
4equals.
[2015]
Options:
A.4l mn2
B.4l 2m2n2
C.4l 2mn
8
5
2+12
5
2+16
5
2+20
5
2... + 44
5
2
S=16
25(22+32+42+ ... + 112)
=16
25
11(11 +1)(22 +1)
61=16
25 ×505 =16
5×101
16
5m=16
5×101
m=101
( ) ( ) ( ) ( ) ( )
( )
S= (1+x)2016 +x(1+x)2015 +x2(1+x)2014
+.... + x2015(1+x) + x2016... (i)
x
1+xS=x(1+x)2015 +x2(1+x)2014 + .....+
x2016 +x2017
1+x... (ii)
Subtracting(i)from(ii)
S
1+x= (1+x)2016 x2017
1+x
S= (1+x)2017 x2017
a17 = coefficientofx17 =2017C17 =2017!
17!2000!
( )
D.4l m2n
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question222
Thesumofthe3 rd andthe4 th termsofaG.P.is60andtheproductofits
firstthreetermsis1000.IfthefirsttermofthisG.P.ispositive,thenits7 th
termis:
[OnlineApril11,2015]
Options:
A.7290
B.640
C.2430
D.320
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question223
Thesumoffirst9termsoftheseries.
13
1+13+23
1+3+13+23+33
1+3+5+ ...
[2015]
Options:
A.142
B.192
C.71
m=l+n
2andcommonratioofG.P.
=r=n
l
1
4
G1=l34n14,G2=l12n12,G3=l14n34
G1
4+2G2
4+G3
4=l3n+2l 2n2+l n3
=ln(l+n)2=ln × (2m)2=4l m2n
( )
Leta,arandar2bethefirstthreetermsofG.PAccordingtothequestion
a(ar)(ar2) = 1000 (ar)3=1000 ar =10
andar2+ar3=60 ar(r+r2) = 60
r2+r6=0
r=2, 3
a=5,a= 10
3reject)
Hence,T7=ar6=5(2)6=5×64 =320
(
D.96
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question224
If 5
n=1
1
n(n+1)(n+2)(n+3)=k
3,thenkisequalto
[OnlineApril11,2015]
Options:
A. 1
6
B. 17
105
C. 55
336
D. 19
112
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question225
Thevalueof
r=16
30 (r+2)(r3)isequalto:
[OnlineApril10,2015]
Options:
A.7770
B.7785
nth termofseries =
n(n+1)
2
2
n2=1
4(n+1)2
Sumofnterm = Σ1
4(n+1)2=1
4[Σn2+2Σn +n]
=1
4
n(n+1)(2n +1)
6+2n(n+1)
2+n
Sumof9terms
=1
4
9×10 ×19
6+18 ×10
2+9=384
4=96
[ ]
[ ]
[ ]
Generaltermofgivenexpressioncanbewrittenas
Tr=1
3
1
n(n+1)(n+2)1
(n+1)(n+2)(n+3)
ontakingsummationboththeside,weget
r=1
5Tr=1
3
1
61
6.7.8 =k
3
1
3×1
611
56 =k
31
3×1
6×55
56 =k
3
k=55
336
[ ]
[ ]
( )
C.7775
D.7780
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question226
Letαandβbetherootsofequationpx2+qx +r=0,p0Ifp,q,rarein
A.Pand 1
α+1
β=4,thenthevalueof|αβ|is:
[2014]
Options:
A. 34
9
B. 213
9
C. 61
9
D. 217
9
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question227
Thesumofthefirst20termscommonbetweentheseries3
+7+11 +15 + .........and1 +6+11 +16 + ......,is
20
r=1
(r2r6) = 7780
Letp,q,rareinAP
2q =p+r...(i)
Given 1
α+1
β=4
α+β
αβ =4
Wehaveα+β= qpandαβ =r
p
xq
p
r
p
=4q= 4r...(ii)
From(i),wehave
2(−4r) = p+r
p= 9r...(iii)
Now,|αβ| = (α+β)24αβ
=q
p
24r
p=q24pr
|p|
From(ii)and(iii)
=16r2+36r2
|−9r|=213
9
( )
[OnlineApril11,2014]
Options:
A.4000
B.4020
C.4200
D.4220
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question228
GivenanA.P.whosetermsareallpositiveintegers.Thesumofitsfirstnine
termsisgreaterthan200andlessthan220.Ifthesecondterminitis12,
thenits4 th termis:
[OnlineApril9,2014]
Options:
A.8
B.16
C.20
D.24
Answer:C
Solution:
Solution:
Givenn=20;S20 = ?
Series(1) 3,7,x11,15,19,23,27,x31,35,39,43,4751,55,59...
Series(2) 1,6,x11,16,21,26,x31,36,41,46,x51,5661,66,71
Thecommontermsbetweenboththeseriesare11,31,51,71...
AboveseriesformsanArithmeticprogression(A.P).
Therefore,firstterm(a) = 11and
commondifference(d) = 20
Now,Sn=n
2[2a + (n1)d]
S20 =20
2[2×11 + (20 1)20]
S20 =10[22 +19 ×20]
S20 =10 ×402 =4020
S20 =4020
LetabethefirsttermanddbethecommondifferenceofgivenA.P.
Secondterm,a+d=12...(i)
Sumoffirstnineterms,
S9=9
2(2a +8d ) = 9(a+4d )
GiventhatS9ismorethan200andlessthan220
200 <S9<220
200 <9(a+4d ) < 220
200 <9(a+d+3d ) < 220
Puttingvalueof(a+d)fromequation(i)
200 <9(12 +3d ) < 220
200 <108 +27d <220
200 108 <108 +27d 108 <220 108
-------------------------------------------------------------------------------------------------
Question229
ThreepositivenumbersformanincreasingG.P.Ifthemiddleterminthis
G.P.isdoubled,thenewnumbersareinA.P.thenthecommonratioofthe
G.P.is:
[2014]
Options:
A.2 3
B.2 + 3
C.2+ 3
D.3 + 2
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question230
Theleastpositiveintegernsuchthat1 2
32
32 .... 2
3n1<1
100 ,is:
[OnlineApril12,2014]
Options:
A.4
B.5
C.6
D.7
Answer:B
Solution:
92 <27d <112
Possiblevalueofdis4
27 ×4=108
Thus,92 <108 <112
Puttingvalueofdinequation(i)
a+d=12
a=12 4=8
4th term = a+3d =8+3×4=20
Leta,ar,ar2areinG.P.
Accordingtothequestion
a,2ar,ar2areinA.P.
2×2ar =a+ar2
4r =1+r2r24r +1=0
r=4± 16 4
2=2± 3
Sincer>1
r=2 3 isrejected
Hence,r=2+ 3
-------------------------------------------------------------------------------------------------
Question231
Inageometricprogression,iftheratioofthesumoffirst5termstothesum
oftheirreciprocalsis49,andthesumofthefirstandthethirdtermis35.
Thenthefirsttermofthisgeometricprogressionis:
[OnlineApril11,2014]
Options:
A.7
B.21
C.28
D.42
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question232
Thecoefficientofx50inthebinomialexpansionof
(1+x)1000 +x(1+x)999 +x2(1+x)998 + ...+x1000is:
12
32
32...⋅ 2
3n1<1
100
12
3
1
3+1
32+1
33+ ... 1
3n1<1
100
121
3
1
3n1
1
31
<1
100
123n1
2.3n<1
100
13n1
3n<1
100
11+1
3n<1
100 100 <3n
Thus,leastvalueofnis5
[ ]
[ ( ) ]
[ ]
[ ]
AccordingtoQuestion
S5
S5
=49 \{here,S5= Sumoffirst5termsandS5= Sumoftheirreciprocals)
a(r51)
(r1)
a1(r51)
(r11)
=49
a(r51) × (r11)
a1(r51) × (r1)=49
or a2(1r5) × (1r) × r5
(1r5) × (1r) × r=49
a2r4=49 a2r4=72
ar2=7...(i)
Also,given,S1+S3=35
a+ar2=35...(ii)
Nowsubstitutingthevalueofeq.(i)ineq.(ii)
a+7=35
a=28
[OnlineApril11,2014]
Options:
A. (1000)!
(50)!(950)!
B. (1000)!
(49)!(951)!
C. (1001)!
(51)!(950)!
D. (1001)!
(50)!(951)!
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question233
LetGbethegeometricmeanoftwopositivenumbersaandb,andM bethe
arithmeticmeanof 1
aand 1
b.If 1
M:Gis4 :5,thena :bcanbe:
[OnlineApril12,2014]
Options:
A.1:4
B.1:2
C.2:3
D.3:4
Answer:A
Solution:
Solution:
Letgivenexpansionbe
S= (1+x)1000 +x(1+x)999 +x2(1+x)998 + ...+... + x1000
Put1+x=t
S=t1000 +xt999 +x2(t)998 + ... + x1000
ThisisaG.Pwithcommonratio x
t
S=
t1000 1x
t
1001
1x
t
=
(1+x)1000 1x
1+x
1001
1x
1+x
=(1+x)1001[(1+x)1001 x1001]
(1+x)1001
= [(1+x)1001 x1001]
Nowcoeffofx50inaboveexpansionisequaltocoeffofx50in(1+x)1001whichis1001C50
=(1001)!
50!(951)!
[ ( ) ]
[ ( ) ]
G= ab
M=
1
a+1
b
2
-------------------------------------------------------------------------------------------------
Question234
If(10)9+2(11)1(108) + 3(11)2(10)7+ ....+10(11)9=k(10)9,thenkisequalto:
[2014]
Options:
A.100
B.110
C. 121
10
D. 441
100
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question235
M=a+b
2ab
Giventhat 1
M:G=4:5
2ab
(a+b)√ab =4
5
a+b
2ab =5
4
a+b+2ab
a+b2ab =5+4
54
{UsingComponendo&Dividendo}
(√a)2+ (√b)2+2ab
(√a)2+ (√b)22ab =9
1
b+ a
b a
2
=9
1b+ a
b a=3
1
b+ a+ b a
b+ a b+ a=3+1
31
{UsingComponendo&Dividendo}
b
a=4
2=2
b
a=4
1
a
b=1
4a:b=1:4
( )
Giventhat109+2 (11)(10)8+3(11)2(10)7+ ... + 10(11)9 = k(10)9
Letx=109+2 (11)(10)8+3(11)2(10)7+ ... + 10(11)9...(i)
Multipliedby 11
10onboththesides
11
10x=11.108+2. (11)2 (10)7+ ... + 9(11)9+1110...(ii)
Subtract(ii)from(i),weget
x 1 11
10 =109+11(10)8+112× (10)7+ ... + 1191110
x
10 =109
11
10
10 1
11
10 1
1110
x
10 = (1110 1010) 1110 = 1010
x=1011 =k109Given
k=100
( )
[( ) ]
ThenumberoftermsinanA.P.iseven;thesumoftheoddtermsinitis24
andthattheeventermsis30.Ifthelasttermexceedsthefirsttermby10 1
2,
thenthenumberoftermsintheA.P.is:
[OnlineApril19,2014
Options:
A.4
B.8
C.12
D.16
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question236
Ifthesum 3
12+5
12+22+7
12+22+32+ ......+upto20termsisequalto k
21 ,thenkis
equalto:
[OnlineApril9,2014]
Options:
A.120
B.180
C.240
D.60
Answer:A
Solution:
Leta,dand2nbethefirstterm,commondifferenceandtotalnumberoftermsofanA.P.respectivelyi.e.a
+(a+d) + (a+2d ) + ... + (a+ (2n 1)d)
No.ofeventerms = n,No.ofoddterms = n
Sumofoddterms:
So=n
2[2a + (n1)(2d )] = 24
n[a+ (n1)d] = 24...(i)
Sumofeventerms:
Se=n
2[2(a+d) + (n1)2d ] = 30
n[a+d+ (n1)d] = 30...(ii)
Subtractingequation(i)from(ii),weget
nd =6...(iii)
Also,giventhatlasttermexceedsthefirsttermby 21
2
a+ (2n 1)d=a+21
2
2nd d=21
2
2×621
2=d(∵nd =6)
d=3
2
Puttingvalueofdinequation(3)n=6×2
3=4
Totalno.ofterms = 2n =2×4=8
Solution:
-------------------------------------------------------------------------------------------------
Question237
Leta1,a2,a3, ...beanA.P,suchthat a1+a2+ ... + ap
a1+a2+a3+ ... + aq
=p3
q3;pq.Then a6
a21
is
equalto:
[OnlineApril9,2013]
Options:
A. 41
11
B. 31
121
C. 11
41
D. 121
1861
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question238
Thesumoftheseries:1 +1
1+2+1
1+2+3+ .......upto10terms,is
[OnlineApril9,2013]
Options:
nth termofgivenseriesis
2n +1
n(n+1)(2n +1)
6
=6
n(n+1)
Letnth term,an=61
n1
n+1
Sumof20terms,S20 =a1+a2+a3+ .... + a20
S20 =61
11
2+61
21
3+61
31
4+ ...
+61
18 1
19 +61
19 1
20 +61
20 1
21
S20 =11
2+1
21
3+1
31
4+ ...
+1
18 1
19 +1
19 1
20 +1
20 1
21
S20 =6 1 1
21 =120
21 ...(i)
GiventhatS20 =k
21...(ii)
Oncomparing(i)and(ii),weget
k=120
[ ]
( ) ( ) ( )
( ) ( ) ( )
[ ( ) ( ) ( )
( ) ( ) ( ) ]
( )
a1+a2+a3+ ...... + ap
a1+a2+a3+ ...... + aq
=p3
q3
a1+a2
a1
=8
1a1+ (a1+d) = 8a1
d=6a1
Now a6
a21
=a1+5d
a1+20d
=a1+5×6a1
a1+20 ×6a1
=1+30
1+120 =31
121
A. 18
11
B. 22
13
C. 20
11
D. 16
9
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question239
Ifa1,a2,a3, ..., an, ....areinA.P.suchthata4a7+a10 =m,thenthesumof
first13termsofthisA.P.,is:
[OnlineApril23,2013]
Options:
A.10m
B.12m
C.13m
D.15m
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question240
GivensumofthefirstntermsofanA.P.is2n +3n2.AnotherA.P.isformed
withthesamefirsttermanddoubleofthecommondifference,thesumofn
termsofthenewA.P.is:
[OnlineApril22,2013]
Options:
Tr=1
1+2+3+ ... + r=2
r(r+1)S10 =2
10
r=1
1
r(r+1)=2
10
r=1
r+1
r(r+1)r
r(r+1)
=2
10
r=1
1
r1
r+1
=21
11
2+1
21
3+1
31
4+ ... + 1
10 1
11
=2 1 1
11 =2×10
11 =20
11
[ ]
( )
[ ( ) ( ) ( ) ( ) ]
[ ]
Ifdbethecommondifference,then
m=a4a7+a10 =a4a7+a7+3d =a7
S13 =13
2[a1+a13] = 13
2[a1+a7+6d ]
=13
2[2a7] = 13a7=13m
A.n +4n2
B.6n2n
C.n2+4n
D.3n +2n2
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question241
Givenasequenceof4numbers,firstthreeofwhichareinG.P.andthelast
threeareinA.P.withcommondifferencesix.Iffirstandlasttermsofthis
sequenceareequal,thenthelasttermis:
[OnlineApril25,2013]
Options:
A.16
B.8
C.4
D.2
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question242
Thesumoffirst20termsofthesequence0.7,0.77,0.777, .....is
[2013]
Options:
GivenSn=2n +3n2
Now,firstterm = 2+3=5
secondterm = 2(2) + 3(4) = 16
thirdterm = 2(3) + 3(9) = 33
Now,sumgiveninoption(b)onlyhasthesamefirsttermanddifferencebetween2nd and1sttermisdoublealso.
Leta,b,c,dbefournumbersofthesequence.
Now,accordingtothequestionb2=acandcb=6anda
c=6
Also,givena=d
b2=ac b2=aa+b
2(∵2c =a+b)
a22b2+ab =0
Now,cb=6andac=6
givesab=12
b=a12
a22b2+ab =0
a22(a12)2+a(a12) = 0
a22a2288 +48a +a212a =0
36a =288 a=8
Hence,lasttermisd=a=8.
[ ]
A. 7
81(179 1020)
B. 7
9(99 1020)
C. 7
81(179 +1020)
D. 7
9(99 +1020)
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question243
Thevalueof12+32+52+ .................... + 252is
[OnlineApril25,2013]
Options:
A.2925
B.1469
C.1728
D.1456
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
LetS=7
10 +77
100 +777
103+ ......+upto20terms
=71
10 +11
100 +111
103+ ......+upto20terms]
Multiplyanddivideby9
=7
9
9
10 +99
100 +999
1000 + .....upto20terms]
=7
9
11
10 +11
102+11
103
+...... upto 20 terms
=7
920
1
10 11
10
20
11
10
=7
9
179
9+1
9
1
10
20 =7
81[179 + (10)20]
[
[
[( ) ( ) ( ) ]
[( ( ) ) ]
[ ( ) ]
Consider12+32+52+ ...... + 252
nth termTn= (2n 1)2,n=1, ..... . 13
Now, Sn=
13
n=1
Tn=
13
n=1
(2n 1)2
=
13
n=1
4n2+
13
n=1
1
13
n=1
4n =4n2+13 4n
=4n(n+1)(2n +1)
6+13 4n(n+1)
2
Putn=13,weget
Sn=26 ×14 ×9+13 26 ×14
=3276 +13 364 =2925
[ ]
Question244
Thesumoftheseries:
(b)2+2(d)2+3(6)2+ ... upto 10 termsis:
[OnlineApril23,2013]
Options:
A.11300
B.11200
C.12100
D.12300
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question245
Thesum 3
12+5
12+22+7
12+22+32+ ....upto11-termsis:
[OnlineApril22,2013]
Options:
A. 7
2
B. 11
4
C. 11
2
D. 60
11
Answer:C
Solution:
Solution:
22+2(4)2+3(6)2+ ......upto10terms
=22[13+23+33+ ......upto10terms]
=410 ×11
2
2=12100
( )
Givensumis
3
12 +5
12+22+7
12+22+32+ ....
nthterm = Tn
=
2n +1
n(n+1)(2n +1)
6=6
n(n+1)
orTn=61
n1
n+1
Sn= Tn=61
n61
n+1=6n
n6
n+1
=66
n+1=6n
n+1
So,sumupto11termsmeans
S11 =6×11
11 +1=66
12 =33
6=11
2
[ ]
-------------------------------------------------------------------------------------------------
Question246
Thesumoftheseries12+2.22+32+2.42+52+2.62+ .... + 2(2m)2is
[OnlineMay7,2012]
Options:
A.m(2m +1)2
B.m2(m+2)
C.m2(2m +1)
D.m(m+2)2
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question247
ThedifferencebetweenthefourthtermandthefirsttermofaGeometrical
Progresssionis52.Ifthesumofitsfirstthreetermsis26,thenthesumof
thefirstsixtermsoftheprogressionis
[OnlineMay7,2012]
Options:
A.63
B.189
C.728
D.364
Answer:C
Solution:
Solution:
Thesumofthegivenseries12+2.22+32+2.42+52
+2.62+ ........ + 2(2m)2is 2m(2m +1)2
2=m(2m +1)2
Leta,ar,ar2,ar3,ar4,ar5besixtermsofaG.P.where'a'isfirsttermandriscommonratio.
Accordingtogivenconditions,wehave
ar3a=5a(r31) = 52...(i)
anda+ar +ar2=26
a(1+r+r2) = 26...(ii)
Tofind:a(1+r+r2+r3+r4+r5)
Consider
a[1+r+r2+r3+r4+r5]
=a[1+r+r2+r3(1+r+r2)]
=a[1+r+r2][1+r3]...(iii)
Divide(i)by(ii),weget
r31
1+r+r2=2
weknowr31= (r1)(1+r+r2)
r1=2r=3and a=2
a(1+r+r2+r3+r4+r5) = a(1+r+r2)(1+r3)
-------------------------------------------------------------------------------------------------
Question248
Ifa,b,c,d andparedistinctrealnumberssuchthat
(a2+b2+c2)p22p(ab +bc +cd ) + (b2+c2+d2) 0then
[OnlineMay12,2012]
Options:
A.a,b,c,d areinA.P.
B.ab =cd
C.ac =bd
D.a,b,c,d areinG.P.
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question249
Thesumoftheseries 1
1+ 2+1
2+ 3+1
3+ 4+ .upto15termsis
[OnlineMay12,2012]
Options:
A.1
B.2
C.3
D.4
Answer:C
Solution:
Solution:
=2(1+3+9)(1+27) = 26 ×28 =728
Thegivenrelationcanbewrittenas
(a2p22abp +b2) + (b2p2+c22bpc)+(c2p2+d22pcd ) 0
or (ap b)2+ (bp c)2+ (cp d)20...(i)
Sincea,b,c,dandpareallreal,theinequality(i)ispossibleonlywheneachoffactoriszero.
i.e.,ap b=0,bp c=0andcp d=0
or p =b
a=c
b=d
c
ora,b,c,dareinG.P.
Givenseriesis 1
1+ 2+1
2+ 3+1
3+ 4+ ....
nth term = 1
n+ n+1
15 th term =1
15 + 16
Thus,givenseriesupto15termsis
1
1+ 2+1
2+ 3+1
3+ 4+ ...... + 1
15 + 16
Thiscanbere-writtenas
-------------------------------------------------------------------------------------------------
Question250
Supposeθandφ(≠0)aresuchthatsec(θ+φ), sec θandsec(θφ)areinA.P.
Ifcos θ =k cos φ
2forsomek,thenkisequalto
[OnlineMay19,2012]
Options:
A.±2
B.±1
C.±1
2
D.±2
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question251
Thesumoftheseries1 +4
3+10
9+28
27 + ...uptontermsis
[OnlineMay19,2012]
Options:
A. 7
6n+1
62
3.2n1
B. 5
3n7
6+1
2.3n1
C.n +1
21
2.3n
D.n 1
31
3.2n1
Answer:C
( )
1 2
1+2 3
1+3 4
1+ ..... + 15 16
1
(Byrationalization)
= 1+ 2 2+ 3 3+ 4+ ..... 14 + 15
= 1+ 16 = 1+4=3
Hence,therequiredsum = 3
Since,sec(θφ), sec θandsec(θ+φ)areinA.P.,∴2 sec θ =sec(θφ) + sec(θ+φ)
2
cos θ =cos(θ+φ) + cos(θφ)
cos(θφ)cos(θ+φ)
2(cos2θsin2φ) = cos θ[2 cos θ cos φ]
cos2θ(1cos φ) = sin2φ=1cos2φ
cos2θ=1+cos φ =2cos2φ
2
cos θ = 2 cos φ
2
Butgivencos θ =k cos φ
2
k= 2
Solution:
:
-------------------------------------------------------------------------------------------------
Question252
IftheA.M.betweenp th andq th termsofanA.P.isequaltotheA.M.
betweenr th ands th termsofthesameA.P.thenp +qisequalto
[OnlineMay26,2012]
Options:
A.r +s1
B.r +s2
C.r +s+1
D.r +s
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question253
If100timesthe100 th termofanAPwithnonzerocommondifference
equalsthe50timesits50 th term,thenthe150 th termofthisAPis:
[2012]
Options:
A.-150
B.150timesits50 th term
C.150
D.Zero
Answer:D
Givenseriesis1+4
3+10
9+28
27 + ... . nterms
=1+1+1
3+1+1
9+1+1
27 + ... . nterms
= ( 1+1+1+ .... + nterms)
+1
3+1
9+1
27 + ... . n terms
=n+
1
311
3n
11
3
=n+1
3×3
2[13n]
=n+1
2[13n] = n+1
21
2.3n
( ) ( ) ( )
( )
( )
Given: ap+aq
2=ar+aS
2
a+ (p1)d+a+ (q1)d
=a+ (r1)d+a+ (s1)d
2a + (p+q)d2d =2a + (r+s)d2d
(p+q)d= (r+s)dp+q=r+s
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question254
Statement-1:Thesumoftheseries1 + (1+2+4)+
(4+6+9) + (9+12 +16) + .... + (361 +380 +400)is8000
Statement-2: n
k=1(k3 (k1)3) = n3,foranynaturalnumbern.
[2012]
Options:
A.Statement-1isfalse,Statement-2istrue.
B.Statement-1istrue,statement-2istrue;statement-2isacorrectexplanationforStatement-1.
C.Statement-1istrue,statement-2istrue;statement-2isnotacorrectexplanationforStatement-1
.
D.Statement-1istrue,statement-2isfalse.
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question255
Letanbethen th termofanA.P.If 100
r=1a2r =αand 100
r=1a2r 1=β,thenthe
commondifferenceoftheA.P.is
[2011]
Options:
A.α β
B. αβ
100
C.β α
D. αβ
200
Answer:B
Let'aisthefirsttermand'd'isthecommondifferenceof anA .P.
Now,Accordingtothequestion100a100 =50a50
100(a+99d ) = 50(a+49d )
2a +198d =a+49d a+149d =0
Hence,T150 =a+149d =0
nthtermofthegivenseries
=Tn= (n1)2+ (n1)n+n2
=((n1)3n3)
(n1) n=n3 (n1)3
Sn=
n
k=1
[k3 (k1)3] 8000 =n3
n=20whichisanaturalnumber.
Hence,boththegivenstatementsaretrue.
andstatement2iscorrectexplanationforstatement1.
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question256
Ifthesumoftheseries12+2.22+32+2.42+52+ .. . 2.62+ ...uptonterms,
whenniseven,is n(n+1)2
2,thenthesumoftheseries,whennisodd,is
[OnlineMay26,2012]
Options:
A.n2(n+1)
B. n2(n1)
2
C. n2(n+1)
2
D.n2(n1)
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question257
Amansaves₹200ineachofthefirstthreemonthsofhisservice.Ineachof
thesubsequentmonthshissavingincreasesby₹40morethanthesavingof
immediatelypreviousmonth.Histotalsavingfromthestartofservicewillbe
₹11040after
[2011]
Options:
A.19months
B.20months
C.21months
D.18months
LetA.P.bea,a+d,a+2d , .........
a2+a4+ .......... + a200 =α
100
2[2(a+d) + (100 1)2d ] = α.. . (i)
anda1+a3+a5+ ........ + a199 =β
100
2[2a + (100 1)2d ] = β...(ii)
Subtracting(ii)from(i),weget
d=αβ
100
Ifnisodd,therequiredsumis
12+2.22+32+2.42+ ..... + 2(n1)2+n2
=(n1)(n1+1)2
2+n2( n1iseven)
=n1
2+1 n2=n2(n+1)
2
( )
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question258
Apersonistocount4500currencynotes.Letandenotethenumberofnotes
hecountsinthen th minute.Ifa1=a2= ... = a10 =150anda10,a11, ...arein
anAPwithcommondifference-2,thenthetimetakenbyhimtocountall
notesis
[2010]
Options:
A.34minutes
B.125minutes
C.135minutes
D.24minutes
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question259
Thesumtoinfinitetermoftheseries1 +2
3+6
32+10
33+14
34+ sis
[2009]
Options:
A.3
B.4
C.6
D.2
Letnumberofmonths = n
200 ×3+ ( 240 +280 +320 + ... + (n3)th term) = 11040
n3
2[2×240 + (n4) × 40] = 11040 600
(n3)[240 +20n 80] = 10440
(n3)(20n +160) = 10440
(n3)(n+8) = 522
n2+5n 546 =0
(n+26)(n21) = 0
n=21
Till10 th minutenumberofcountednotes = 1500
Remainingnotes = 4500 1500 =3000
3000 =n
2[2×148 + (n1)(−2)] = n[148 n+1]
n2149n +3000 =0
n=125,24
Butn=125isnotpossible
∴Totaltime = 24 +10 =34minutes.
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question260
Thefirsttwotermsofageometricprogressionaddupto12.thesumofthe
thirdandthefourthtermsis48.Ifthetermsofthegeometricprogression
arealternatelypositiveandnegative,thenthefirsttermis
[2008]
Options:
A.-4
B.-12
C.12
D.4
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question261
Inageometricprogressionconsistingofpositiveterms,eachtermequalsthe
sumofthenexttwoterms.Thenthecommonratioofitsprogressionis
equals
[2007]
Options:
A.5
LetS=1+2
3+6
32+10
33+14
34+ ..... . ...(i)
Multiplyingbothsidesby 1
3,weget
1
3S=1
3+2
32+6
33+10
34+ ....... . ...(ii)
Subtractingeqn.(ii)fromeqn.(i),weget
2
3S=1+1
3+4
32+4
33+4
34+ ........ .
2
3S=4
3+4
32+4
33+4
34+ ........ .
2
3S=
4
3
11
3
=4
3×3
2S=3
AT Q
a+ar =12
ar2+ar3=48
ar2(1+r)
a(1+r)=48
12 r2=4, r= 2
( ∵termsarealternately+veand−ve )
a= 12
B.1
2(√51)
C. 1
2(1 5)
D.1
25
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question262
Thesumofseries 1
2!1
3!+1
4! ......uptoinfinityis
[2007]
Options:
A.e
1
2
B.e
+1
2
C.e2
D.e1
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question263
Leta1,a2,a3............betermsonA.P.If a1+a2+ ......... . ap
a1+a2+ ........... + aq
=p2
q2,pq,then a6
a21
equals
[2006]
Options:
A. 41
11
Lettheseriesa,ar,ar2, ....areingeometricprogression.
Giventhat,a=ar +ar2
1=r+r2r2+r1=0
r=1± 14×−1
2=1± 5
2
r=51
2[ ∵termsofG.P.arepositive∴rshouldbepositive]
Weknowthatex=1+x+x2
2!+x3
3!+ ........ .
Putx = 1
e1=11+1
2!1
3!+1
4!...... .
e1=1
2!1
3!+1
4!1
5!........ .
B. 7
2
C. 2
7
D. 11
41
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question264
Thevalueof 10
k=1sin 2kπ
11 +i cos 2kπ
11 is
[2006]
Options:
A.i
B.1
C.-1
D.i
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question265
( )
Giventhat
Sp
Sq
=
p
2[2a1+ (p1)d]
q
2[2a1+ (q1)d]
=p2
q2
2a1+ (p1)d
2a1+ (q1)d=p
q
Putp=11andq=41
a1+5d
a1+20d =11
41 a6
a21
=11
41
10
k=1
sin 2kπ
11 +i cos 2kπ
11
=i
10
k=1
cos 2kπ
11 i sin 2kπ
11 [∵e =cos θ +i sin θ]
=i
10
k=1
e
2kπ
11 i=i
10
k=0
e
2kπ
11 i1
=i 1 +e
11i+e
11i+ .. . .11terms] i
=i1e
11
11
1e
11i
i=i1e2πi
1e
11i
i
=i×0i[∵e2πi =1]
= i
( )
( )
{ }
[
[( ) ][ ]
Iftheexpansioninpowersofxofthefunction 1
(1ax)(1bx)is
a0+a1x+a2x2+a3x3......thenanis
[2006]
Options:
A. bnan
ba
B. anbn
ba
C. an+1bn+1
ba
D. bn+1an+1
ba
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question266
Ifa1,a2, ........., anareinH.P.,thentheexpression
a1a2+a2a3+ .......... + an1anisequalto
[2006]
Options:
A.n(a1an)
B.(n1)(a1an)
C.na1an
D.(n1)a1an
Answer:D
Solution:
Solution:
(1ax)1(1bx)1
= (1+ax +a2x2+ ...)(1+bx +b2x2+ ...)
∴Coefficientofxn
xn=bn+abn1+a2bn2+ ...... + an1b+an
{whichisaG.P.withr=a
b
Itssumis =
bn1a
b
n+1
1a
b
=bn+1an+1
baan=bn+1an+1
ba
[ ( ) ] }
a1,a2,a3..... . anareinH.P.
1
a1
,1
a2
,1
a3
... 1
an
areinA.P.
1
a2
1
a1
=1
a3
1
a2
= ......... = 1
an
1
an1
=d(say)
-------------------------------------------------------------------------------------------------
Question267
Ifthecoefficientsofrth,(r+1)th,and(r+2)thtermsinthethebinomial
expansionof(1+y)mareinA.P.,thenmandrsatisfytheequation
[2005]
Options:
A.m2m(4r 1) + 4r22=0
B.m2m(4r +1) + 4r2+2=0
C.m2m(4r +1) + 4r22=0
D.m2m(4r 1) + 4r2+2=0
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question268
Ifx =
n=0an,y=
n=0bn,z=
n=0cnwherea,b,careinA.Pand
|a| < 1, | b| <1, | c| <1thenx,y,zarein
[2005]
Options:
A.G.P.
B.A.P.
C.Arithmetic-GeometricProgression
D.H.P.
Thena1a2=a1a2
d,a2a3=a2a3
d,
.......... . an1an=an1an
d
Addingallequations,weget
a1a2+a2a3+ ......... + an1an
=a1a2
d+a2a3
d+ .... + an1an
d
=1
d[a1a2+a2a3+ .... + an1an] = a1an
d
Also, 1
an
=1
a1
+ (n1)d
a1an
a1an
= (n1)d
a1an
d= (n1)a1an
Whichistherequiredresult.
Coeficientofrth , (r+1)th and(r+2)th termsismCr1,mCrandmCr+1resp.
GiventhatmCr1,mCr,mCr+1areinA.P.
2mCr=mCr1+mCr+1
2=
mCr1
mCr
+
mCr+1
mCr
=r
mr+1+mr
r+1
m2m(4r +1) + 4r22=0
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question269
Thesumoftheseries1 +1
4.2!+1
16.4!+1
64.6!+ ....................adinf.is
[2005]
Options:
A. e1
e
B. e+1
e
C. e1
2e
D. e+1
2e
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question270
LetT rbetherthtermofanA.P.whosefirsttermisaandcommondifference
isd .Ifforsomepositiveintegersm,n,mn,Tm=1
nandT n=1
m,thena d
equals
[2004]
Options:
x=
n=0
an=1
1aa=11
x
y=
n=0
bn=1
1bb=11
y
z=
n=0
cn=1
1cc=11
z
a,b,careinA.P.⇒ 2b =a+c
2 1 1
y=11
x+11
z
2
y=1
x+1
zx,y,zareinH.P.
( )
Weknowthat
ex+ex
2=1+x2
2!+x4
4!+x6
6!............
Puttingx=1
2,weget
1+1
4.2!+1
16.4!+1
64.6!+ ..... . =e
1
2+e
1
2
2
=
e+1
e
2=e+1
2e
A. 1
m+1
n
B.1
C. 1
mn
D.0
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question271
Lettwonumbershavearithmeticmean9andgeometricmean4.Thenthese
numbersaretherootsofthequadraticequation
[2004]
Options:
A.x218x 16 =0
B.x218x +16 =0
C.x2+18x 16 =0
D.x2+18x +16 =0
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question272
Thesumofseries 1
2!+1
4!+1
6!+ .....is
[2004]
Options:
A. (e22)
e
B. (e1)2
2e
Tm=a+ (m1)d=1
n...(i)
Tn=a+ (n1)d=1
m...(ii)
Subtracting(ii)from(i),weget
(mn)d=1
n1
md=1
mn
From(i)a=1
mn ad=0
Lettwonumbersbeaandbthen a+b
2=9
a+b=18and√ab =4ab =16
∴Equationwithrootsaandbis
x2 (a+b)x+ab =0x218x +16 =0
C. (e21)
2e
D. (e21)
2
Answer:B
Solution:
-------------------------------------------------------------------------------------------------
Question273
Thesumofthefirstntermsoftheseries12+2.22+32+2.42+52+2.62+ ...
is n(n+1)2
2whenniseven.Whennisoddthesum
[2004]
Options:
A. n(n+1)
2
2
B. n2(n+1)
2
C. n(n+1)2
4
D. 3n(n+1)
2
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question274
IfSn=n
r=0
1
nCr
andtn=n
r=0
r
nCr
,then tn
Sn
isequalto
[2004]
[ ]
Weknowthat
ex=1+x
1!+x2
2!+x3
3!... .
e=1+1
1!+1
2!+1
3!+ ......
ande1=11
1!+1
2!1
3!+ .......
e+e1=2 1 +1
2!+1
4!+ ...
1
2!+1
4!+1
6!+ ...... = e+e1
21
=e2+12e
2e =(e1)2
2e
[ ]
Ifnisodd,therequiredsumis
12+2.22+32+2.42+ ...... + 2 (n1)2+n2
=(n1)(n1+1)2
2+n2
[ (n1)iseven
∴usinggivenformulaforthesumof(n1)terms.
=n1
2+1 n2=n2(n+1)
2
( )
Options:
A. 2n 1
2
B. 1
2n1
C.n 1
D. 1
2n
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question275
Ifthesumoftherootsofthequadraticequationax2+bx +c=0isequalto
thesumofthesquaresoftheirreciprocals,then a
c,b
aand c
barein
[2003]
Options:
A.Arithmetic-GeometricProgression
B.ArithmeticProgression
C.GeometricProgression
D.HarmonicProgression.
Answer:D
Solution:
Solution:
Sn=1
nC0
+1
nC1
+1
nC2
+ .... + 1
nCn
tn=0
nC0
+1
nC1
+2
nC2
+ .... + n
nCn
tn=n
nCn
+n1
nCn1
+n2
nCn2
+ ... + 0
nC0
Adding(i)and(ii),weget,
2tn= (n)1
nC0
+1
nC1
+ ... 1
nCn
=nSn
nCr=nCnr
tn
Sn
=n
2
[ ]
ax2+bx +c=0,α+β=b
a,αβ =c
a
AT Q,α+β=1
α2+1
β2
α+β=α2+β2
α2β2 b
a=
b2
a22c
a
c2
a2
Onsimplification2a2c=ab2+bc2
2a
b=c
a+b
cDividebothsidebyabc ]
c
a,a
b,b
careinA.P.
[
-------------------------------------------------------------------------------------------------
Question276
Thesumoftheseries 1
1.2 1
2.3 +1
3.4 ............upto∞isequalto
[2003]
Options:
A.loge
4
e
B.2loge2
C.loge21
D.loge2
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question277
If1,log9(31x+2), log3(4.3x1)areinA.P.thenxequals
[2002]
Options:
A.log34
B.1 log34
C.1 log43
D.log43
Answer:B
Solution:
Solution:
( )
a
c,b
a, & c
bareinH.P.
LetS=1
1.2 1
2.3 +1
3.4.............. .
Tn=1
n(n+1)=1
n1
n+1
S=1
11
21
21
3+1
31
41
41
5....
=121
21
3+1
41
5.............. .
log(1+x) = xx2
2+x3
3x4
4+ ..... .
=12[−log(1+1) + 1] = 2 log 2 1=log 4
e
( )
( ) ( ) ( ) ( )
[ ]
[ ]
( )
1,log9(31x+2), log3(4.3x1)areinA.P.
a,b,careinA.Pthenb=a+c
2log9(31x+2) = 1+log3(4.3x1)
-------------------------------------------------------------------------------------------------
Question278
SumofinfinitenumberoftermsofGPis20andsumoftheirsquareis100.
ThecommonratioofGPis
[2002]
Options:
A.5
B.3
5
C.8
5
D.1
5
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question279
FifthtermofaGPis2,thentheproductofits9termsis
[2002]
Options:
A.256
B.512
C.1024
logbqap=p
qlogba
log3(31x+2) = log33+log3(4.3x1)
log3(31x+2) = log3[3(43x1)]
31x+2=3(4.3x1)
3.3x+2=12.3x3Put3x=t
3
t+2=12t 312t25t 3=0
Hencet= 1
3,3
4
3x=3
4as3x ve )
x=log3
3
4orx=log33log34
x=1log34
(
( )
Leta= firsttermofG.P.andr= commonratioofG.P.
ThenG.P.isa,ar,ar2
GivenS=20 a
1r=20
a=20(1r)...(i)
Alsoa2+a2r2+a2r4+ ...to=100
a2
1r2=100 [20(1r)]2
1r2=100 [from(i) ]
400(1r)2
(1r)(1+r)=100 4(1r) = 1+r
1+r=44r 5r =3r=35
D.noneofthese
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question280
1323+3343+ ... + 93=
[2002]
Options:
A.425
B.-425
C.475
D.-475
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question281
Thevalueof2144188116...∞is
[2002]
Options:
A.1
B.2
C.3
2
D.4
Answer:B
Solution:
a4=2ar4=2
Now,a×ar ×ar2×ar3×ar4×ar5×ar6×ar7×ar8
=a9r36 = (ar4)9=29=512
1323+3343+ ....... + 93
=13+23+33+ ...... + 932(23+43+63+83)
Σn3=n(n+1)
2
2
=9×10
2
22.23[13+23+33+43]
= (45)216 4×5
2
2=2025 1600 =425
[ ( ) ]
[ ]
[ ]
-------------------------------------------------------------------------------------------------
Now,letS=1
4+2
8+3
16 + ........ . ..... . (i)
1
2S=1
8+2
16 + ....... . .....(ii)
Subtracting(ii)from(i)
1
2S=1
4+1
8+1
16 + ........ .
or 1
2S=a
1r=14
112=1
2S=1
P=2S=2
LetP=2142282316........ .
=214+28+316 + ........ .